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Interpreting a balanced chemical reaction - class-IX

Description: interpreting a balanced chemical reaction
Number of Questions: 45
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Tags: chemistry introduction to analytical chemistry stoichiometric relationships chemical equations quantitative chemistry
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0.05 mole of $LiAlH _4$ in ether solution was placed in a flask containing 74g (1 mole ) of t-butyl alcohol.The product $LiAlHC _{12}H _{27}O _3$ weighed 12.7 g. If Li atoms are conserved, the percentage yield is : (Li=7, Al=27, H=1, C=12, O=16).

  1. 25%

  2. 75%

  3. 100 %

  4. 15 %


Correct Option: C

$1.25$ g of sample of limestone on heating gives $0.44$ g carbon dioxide. The percentage purity of $CaCO _3$ in limestone is:

  1. $75\%$

  2. $85\%$

  3. $90\%$

  4. $80\%$


Correct Option: D
Explanation:

$CaCO _3$ $\quad \underrightarrow \Delta\quad  CaO +CO _2$

Moles of $CO _2$ produced =$\cfrac{0.44}{44}$$=0.01$ moles
Moles of pure $CaCO$$ _3$ required $=0.01$ moles
$=0.01 \times 100g$
$=1g$
Therefore $\%$ purity of $CaCO$$ _3$  in limestone =$\cfrac{1}{1.25}\times 100$
$=80\%$

$50\ g$ of an impure calcium carbonate sample decomposes on heating to give carbon dioxide and $22.4\ g$ calcium oxide. The percentage purity of calcium carbonate in the sample is:

  1. $60\%$

  2. $80\%$

  3. $90\%$

  4. $70\%$


Correct Option: B
Explanation:

$CaCO _3\overset { \Delta  }{ \rightleftharpoons  } CaO+CO _2$

Let pure sample of $CaCO _3=x$ $grams$
Mass of $CaO$ produced after decomposition $=22.4$ $g$
Molar mass of $CaCO _3=100$ ${g/mol}$
and, Molar mass of $CaO=56$ $g$
If $100\%$ is pure, then
$100$ $g$ $CaCO _3\longrightarrow 56$ $g$ of $CaO$
Also,$y$ $g$ of $CaCO _3\longrightarrow 22.4$ $g$ of $CaO$
Dividing these two,
$\cfrac{100}{y}=\cfrac{56}{22.4}$
$y=40$ $grams$
$\therefore$ Percentage of purity $=\cfrac{Mass\quad of \quad pure\quad sample}{Total\quad mass\quad of\quad impure}\times 100=\cfrac{40}{50}\times 100=80\%$

Statement (A):The constant temperature at which a solid change into liquid state by absorbing heat energy is called melting point.
Statement (B): The constant temperature at which a liquid changes into gaseous state by absorbing heat energy is called point.
Statement (C): The melting and boiling points of a pure substance is always a constant and is a good measure of finding the purity of the substance.
  1. All the statement are correct

  2. All the statement are incorrect

  3. A, B are correct and C is incorrect

  4. A, B are incorrect and C is correct


Correct Option: A
Explanation:

$A\rightarrow $ Its defination of melting point.

$B\rightarrow $ Its defination of boiling point. 
$C\rightarrow $ Pure substance have constant boiling and melting point. 
$\therefore $ All are true.

If the percentage yield of the $1 st$ step is $80\% $ and that of the $2nd $ step is $75\% $, then what is the expected overall percentage yield for producing $CaO _3$ from $CaCl _{2} $?

  1. $50\%$

  2. $70\%$

  3. $55\%$

  4. $60\%$


Correct Option: D
Explanation:
${ CaCl } _{ 2 }\xrightarrow [  ]{ { 1 }^{ st }step } \times \xrightarrow [  ]{ { 2 }^{ nd }step } { CaCO } _{ 3 }$
Let $100$ unit of $CaCl _2$ taken
amount of $\times $ produced $=80\ unit$
amount of $CaCO _3$ produced $=\dfrac {80\times 75}{100}=60$
Hence $\%$ yield of $CaCO _3$ by $CaCl _2=60\%$

What is the purity of concentrated $H _2SO _4$ solution $(d=1.8gm/mol)$ if $5\text{ ml}$ of these solution is neutralized by $84.5 \text{ ml}$ of $2N \text{ NaOH}$ solution. 

  1. $93 \text {%}$

  2. $94.6 \text {%}$

  3. $92.12 \text {%}$

  4. $91.5 \text {%}$


Correct Option: C
Explanation:

$H _2SO _4+2NaOH \rightarrow Na _2SO _4+2H _2O$
Moles of $NaOH$ required $=\cfrac {2 \times 84.5}{1000}=0.169$
For $2 \ moles \ NaOH \rightarrow 1 \ mole \ H _2SO _4$ is used
For $\ 0.169 \ mole \ NaOH \rightarrow 0.0845 \ moles$ are used.
Mass of $H _2SO _4 \rightarrow 0.0845 \times 98=8.281 \ gm$ (Theoretical)
Mass of $H _2SO _4$ used in original reaction $\Rightarrow 5 \times 1.8 = 9 \ gm$
$\therefore$ % purity $=\cfrac {8.281}9 \times 100 = 92.12$ %

For the reaction, $2Fe(NO _3) _3+3Na _2CO _3\rightarrow Fe _2(CO _3) _3+6NaNO _3$ initially 2.5 mole of $Fe(NO _3) _2$ and 3.6 mole of $Na _2CO _3$ are taken. If 6.3 mole of $NaNO _3$ is obtained then % yield of given reaction is:

  1. 50

  2. 84

  3. 87.5

  4. 100


Correct Option: C
Explanation:

$\quad \quad \quad \quad 2Fe{ (N{ O } _{ 3 }) } _{ 3 }+3{ Na } _{ 2 }{ CO } _{ 3 }\longrightarrow { Fe } _{ 2 }{ (C{ O } _{ 3 }) } _{ 3 }+6{ Na }{ NO } _{ 3 }\ initial\quad \quad 2.5\quad \quad \quad \quad \quad 3.6\quad \quad \quad \quad \quad  \quad - \quad \quad \quad \quad -\ after\quad 2.5-2.4=0.1\quad \quad 0\quad \quad \quad \quad \quad \quad 1.2 \quad \quad \quad \quad 6.3\ reaction$

As 2 moles of $Fe(NO _{3}) _{3}$ reacts with 3 moles of $Na _{2}CO _{3}$. 
Thus 2.5 moles of $Fe(NO _{3}) _{3}$ reacts with=$\cfrac{3}{2} \times 2.5$ moles of $Na _{2}CO _{3}$=3.75 moles of $Na _{2}CO _{3}$
As $Na _{2}CO _{3}$ present is 3.6 moles only. Thus, $Na _{2}CO _{3}$ is limiting reagent.
Now, 3 moles of $Na _{2}CO _{3}$ gives=6 moles of $NaNO _{3}$ 
3.6 moles of $Na _{2}CO _{3}$ gives=$\cfrac{6}{3} \times 3.6$ moles of $NaNO _{3}$
3.6 moles of $Na _{2}CO _{3}$ gives=7.2 moles of $NaNO _{3}$
But $NaNO _{3}$ obtained is 6.3 moles.
Thus % yield=$\cfrac{6.3}{7.2} \times 100=87.5 \%$

$12.5$ g of an impure sample of limestone on heating gives $4.4$ g of carbon dioxide. The percentage purity of $CaCO _{3}$ in the sample is:

  1. $72$%

  2. $75$%

  3. $80$%

  4. $85$%


Correct Option: C
Explanation:

${ CaCO } _{ 3 }\overset { \Delta  }{ = } CaO+{ CO } _{ 2 }\uparrow $

$100gm$                      $44gm$
Therefore $1$ mole i.e. $100gm$ of ${ CaCO } _{ 3 }$ (lime stone) give $1$ mole i.e. $44gm$ of ${ CO } _{ 2 }$.
$4.4gm$ of ${ CO } _{ 2 }$ are produced from $\dfrac { 100\times 4.4 }{ 44 } gm$ i.e. $10gm$ of ${ CaCO } _{ 3 }$.
$\therefore$   The percentage of purity $=\left( 1-\dfrac { 12.5-10 }{ 12.5 }  \right) \times 100$% $=80$%
$\therefore$   Correct answer is $C$ $(80$%$)$.

For the complete reduction of $5.8g$ of acetone to isopropyl alcohol, the quantity of $LiAIH _{4}$ required (assuming chemical yield to be $100\%$ ) is approximately [mass:$Li=6.9,Al=27$]

  1. $5.8\ g$

  2. $3.8\ g$

  3. $1.9\ g$

  4. $15.2\ g$


Correct Option: B

0.2828 g of iron wire was dissolved in excess of dilute $H _2SO _4$ and the solution was made upto 100 ml. 20 ml of this solution required 30 ml. of $\dfrac{N}{30} K _2Cr _2O _7$ solution for oxidation. Calculate % purity of iron in the wire:

  1. 99

  2. 95

  3. 90

  4. 85


Correct Option: B

To a $10$ml $1M$ aqueous solution of $Br _2$,excess of NaOH is added so that all $Br _2$ is disproportional to $Br^-$ and $BrO _3^-$, the resulting solution is freed from $Br^-$,by extraction and excess of $OH^-$ neutralised by acidifying the solution. The resulting solution is sufficient the react with $1.5$gm of impure $CaC _2I _4$ $(M=128gm /mol)$ sample. The purity by mass of Oxalate sample is the relevant reaction s are $Br _2(aq.)+OH^- \rightarrow (aq.)+BrO _3^-$
$Bro _3^-+C _2O _4^{2-}\rightarrow Br^-+CO _2$

  1. $85.3\%$

  2. $12.5\%$

  3. $90\%$

  4. $50\%$


Correct Option: A

In the decomposition of 10 g of $Mg{ CO } _{ 3 }$, 0.1 mole ${ CO } _{ 2 }$ and 4.0 g MgO are obtained. Hence, percentage purity of $Mg{ CO } _{ 3 }$ is:

  1. 50%

  2. 60%

  3. 40%

  4. 84%


Correct Option: D
Explanation:

Solution:- (D) $84 \%$

Molecular weight of $MgC{O} _{3} = 84 \; g$
Molecular weight of $MgO = 40 \; g$

Decomposition of $MgC{O} _{3}$-
$MgC{O} _{3} \longrightarrow MgO + C{O} _{2}$

Now, from the above reaction-
Weight of pure $MgC{O} _{3}$ required to produce $40 \; g$ of $MgO = 84.3 \; g$
Weight of pure $MgC{O} _{3}$ required to produce $4 \; g$ of $MgO = \cfrac{84.3}{40} \times 4 = 8.43 \; g$
Given weight of $MgC{O} _{3} = 10 \; g$
Now,
Amount of pure $MgC{O} _{3}$ in $10 \; g$ of given $MgC{O} _{3} = 8.43 \; g$
Thus,
Amount of pure $MgC{O} _{3}$ in $100 \; g$ of given $MgC{O} _{3} = \cfrac{8.43}{10} \times 100 = 84.3 \; g$
Therefore,
The percentage purity of given $MgC{O} _{3} = 84.3 \% \approx 84 \%$

$10g$ of limestone on heating produces $4.2g$ of $CaO$. the percentage purity of $Ca{ CO } _{ 3 }$ in limestone is: 

[Atomic mass of $Ca =$ $40$]

  1. $85%$

  2. $75%$

  3. $95%$

  4. $80%$


Correct Option: B
Explanation:

$10g$ of limestone i.e. $CaCO _3$ contains= $\cfrac {10g}{100g/mole}$ moles of $CaCO _3=0.1$ moles of $CaCO _3$

$CaCO _3 \longrightarrow CaO+CO _2$
$1$ mole of $CaCO _3$ produce $1$ mole of $CaO$
Thus $0.1$ moles of $CaCO _3$ must produce $0.1$ mole of $CaO$
$10g$ of $CaCO _3$ must produce $0.1 \times 56= 5.6g$ of $CaO$
But $CaO$ produce is $4.2g$
Pure product obtained is $4.2g$ from $10g$ of $CaCO _3$
Product that obtain along with $1$ m purity from $10g$ of $CaCO _3$ is $5.6g$
So, percentage purity= $\cfrac {\text {mass of pure substance obtained}}{\text {mass of impure substance obtained}}\times 100$
% purity= $\cfrac {4.2}{5.6}\times 100= 75$%

Which of the following is not the criterion of purity of a substance?

  1. Solubility

  2. Melting point

  3. Boiling point

  4. Density


Correct Option: A,D
Explanation:

A pure substance shows a sharp melting point & boiling point. Hence melting point & boiling point determines the purity.

But density & solubility are not the criterion of purity of substance.

Consider the following reaction sequence${ CaCl } _{ 2(aq) }\quad +\quad { CO } _{ 2(g) }\quad +\quad { H } _{ 2 }O\rightarrow { CaCO } _{ 3(s) }\quad +\quad { 2HCl } _{ (aq) }$${ CaCO } _{ 3(s) }\quad \xrightarrow { heat } { CaO } _{ (s) }\quad +\quad { H } _{ 2 }{ O } _{ (g) }$if the percentage yield of the $1st$ step is $80%$ and that of the $2nd$ is $75%$, then what is the expected overall percentage yield producing $CaO$ from ${ CaCl } _{ 2 }$?

  1. $50%$

  2. $70%$

  3. $55%$

  4. $60%$


Correct Option: D
Explanation:
${ CaCl } _{ 2 }\longrightarrow { CaCO } _{ 3 }$            from question
$100gm\longrightarrow 80gm$
${ CaCO } _{ 3 }\longrightarrow CaO$
$100gm\longrightarrow 75gm$
$80$% $\longrightarrow 60$%
$\therefore$   The percentage of yield of $CaO$ is $60$%.

Methyl benzoate is prepared by the reaction between benzoic acid and methanol, according to the equation 
$ C _{ 6 }H _{ 5 }COOH+CH _{ 3 }OH\longrightarrow C _{ 6 }H _{ 5 }COOCH _{ 3 }+H _{ 2 }O$
Benzoic acid     Methanol           Methyl benzoate
In an experiment 24.4 gm of benzoic acid were reacted with 70.0 mL of $ CH _3OH $. The density of $ CH _3OH $ is $ 0.79 g mL^{-1} $. The methyl benzoate produced had a mass of 21.6g. What was the percentage yield of product ?

  1. 91.7%

  2. 79.4%

  3. 71.5%

  4. 21.7%


Correct Option: A

A sample contain $Fe\left (SO _{4}  \right ) _{3},$ $FeSO _{4a}$ and impurities. A 600 g sample contains 48g impurities ans equal moles of $Fe _{2}\left (SO _{4}.  \right )in the % of Fe _{2}\left ( SO _{4} \right ) _{3}$ in the mixture is:

  1. 33.33%

  2. 66.7%

  3. 83.33%

  4. none of these


Correct Option: B

Percentage purity of a sample of gold is $$. How many atoms of gold are present in its $1$ gram
(Atomic mass of gold =$197 u.) 

  1. $2.6*{10^{21}}$

  2. $2.6*{10^{23}}$

  3. $3.0*{10^{21}}$

  4. $4.5*{10^{20}}$


Correct Option: A

For the reaction :

     
$CaO + 2HCl \to CaC{l _2} + {H _2}O$

$2.46 g$ of CaO is reacted with excess of HCl and $3.7 g$ $ CaC{l _2}$ is formed. What is  percentage yield? 

[$Note :  \%\  Yield = \dfrac{{Actual\,yield}}{{Theoretical\,yield}} \times 100$] 

  1. 86%

  2. 26%

  3. 76%

  4. 16%


Correct Option: C

$4$ g of hydrogen $(H _2)$, $64$g of sulphur (S) and $44.8$ L of $O _2$ at STP react and form $H _2SO _4$. If $49$g of $H _2SO _4$ is formed, then $\%$ yield is  ?

  1. $25\%$

  2. $50\%$

  3. $75\%$

  4. $100\%$


Correct Option: B
Explanation:

$H _2+S+2O _2\longrightarrow H _2SO _4$

 $2g$    $32g$    $\underset {|||}{64g}$           $98g$
                  $44.8L$
The mole ratio is $H:S:O=1:1:2$
Since $O _2$ is limiting only $98g$ of $H _2SO _4$ is formed.
Given $49g$ of $H _2SO _4$ is formed.
So % yield= $\cfrac {49}{98}\times 100=50$% .

With the amounts of reactants provided, it was possible to produce $0.667\ g$ of aspirin. One student produces $0.333\ g$ of aspirin. What was the percent yield for this student's laboratory work?

  1. $40$%

  2. $33$%

  3. $67$%

  4. $50$%


Correct Option: D
Explanation:

Percentage Yield of a compound is defined as ratio of actual yield to the theoretical yiald.

Actual Yield $(E) = 0.333 \space g$
Theoretical Yield $(T) = 0.667 \space g$
$\Rightarrow \% $ Yield $\dfrac{0.333}{0.667} \times 100 = 50\%$

A decomposition reaction produces sodium carbonate from sodium bicarbonate.
If the collected mass of sodium carbonate was $3.7\ g$ and the predicted amount was $4.0\ g$, what is the percent yield of the reaction?

  1. $92.5\%$

  2. $95\%$

  3. $7.5\%$

  4. $90\%$


Correct Option: A
Explanation:

Percent Yield of a compound is defined as ratio of actual yield to the theoretical yield.

$\Rightarrow$ Actual Yield $ = 3.7 \space g$
Theoretical Yield $ = 4.0 \space g$
So, $\% $ Yield $\dfrac{3.7}{4} \times 100$$= 92.5\%$
So, Percent Yield $= 92.5\%$

A metal oxide (MO) is reduced by heating it in a stream of hydrogen. It is found that after complete reduction, 7.95 g of oxide requires 0.2 g of $H _2$ to yield 6.35 g of the metal. We may deduce that:

  1. The atomic weight of the metal is 48

  2. The atomic weight of the metal is 16

  3. The atomic weight of the metal is 12

  4. The atomic weight of the metal is 63.5


Correct Option: D
Explanation:

$2MO + 2H _2 \rightarrow 2M + 2H _2O$

7.92 g    0.2g       6.35 g
let metal weight is x
mol  mol conclution 
$\dfrac{0.2}{2}$ = $\dfrac{6.35}{x}$
we get      x  =  63.5  
ans is D

20 g of a magnesium carbonate sample decomposes on heating to given carbondioxide, and 8g magnesium oxide. What will be the percentage of purity of ${\text{MgC}}{{\text{O}} _3}$ sample ? 

  1. 96

  2. 60

  3. 84

  4. 75


Correct Option: C

How is percent yield calculated?

  1.  $\displaystyle   \text { Percent yield } = \dfrac { \text { actual yield } }{ \text { theoretical yield } } \times  100 $ 

  2.  $\displaystyle   \text { Percent yield } = \dfrac { \text { actual yield } }{ \text { theoretical yield }  \times  100 }  $ 

  3.  $\displaystyle   \text { Percent yield } = \dfrac { \text { actual yield } }{ \text { theoretical yield } } $

  4.  $\displaystyle   \text { Percent yield } = \dfrac {  \text { theoretical yield } }{ \text { actual yield } } $

  5.  $\displaystyle   \text { Percent yield } =  \text { actual yield } \times \text { theoretical yield  } \times  100 $


Correct Option: A
Explanation:

$\displaystyle   \text { Percent yield } = \dfrac { \text { Actual yield } }{ \text { Theoretical yield } } \times  100 $ %

For example,
If actual yield is 10 g and theoretical yield is 20 g, 


Then, percent yield will be-$\displaystyle  \dfrac {10 g}{20g}  \times 100 = 50$ %.


Hence, option $A$ is correct.

What equation is used to calculate percent yield?
Note: $E=$ experimental; $T=$ theoretical

  1. $\cfrac{E}{T}\times 100$

  2. $\cfrac{(E-T)}{T}\times 100$

  3. $\cfrac{T}{E}\times 100$

  4. $\cfrac{(T-E)}{T}\times 100$


Correct Option: A
Explanation:

The calculated or expected amount of product is called theoretical yield. The amount of product actually produced is called actual yield.

When we divide actual yield by the theoretical yield and then multiplied by 100 to get the percentage yield of reaction.
$\Rightarrow \dfrac{E}{T} \times 100 = \% $ Yield

$^{14} _6C\rightarrow ^{14} _7N+X$
Water is formed by the addition of 4.0g of $H _2(g)$ to an excess of $O _2(g)$. If 27 g of $H _2O$ is recovered, what is the percent yield for the reaction?

  1. 25%

  2. 50%

  3. 75%

  4. 100%

  5. Cannot be determined


Correct Option: C
Explanation:

$2H _2 + O _2 \rightarrow 2H _2O$

4 g   excess   2 mol 

then water is also formed 2 mol  =  36 gram 
but it formed only 27 gram

% yeald = $\dfrac{27}{36}\times 100$ 
=  $75%$
ans is C

Consider the reaction:
$2ZnS + 3O _{2}\rightarrow 2ZnO + 2SO _{2}$
This reaction has an $80.0$ yield.
What mass of $ZnO$ is produced when $50.0\ g\ ZnS$ is heated in an open vessel untill no further weight loss is observed?

  1. $33.4\ g$

  2. $40.4\ g$

  3. $43.4\ g$

  4. $3240\ g$


Correct Option: A
Explanation:

Molar mass of $ZnS = 97.5\space g$

No. of moles of $ZnO = \dfrac{50}{97.5} = 0.5128\space moles$
$2\space moles$ of $ZnS$ produce $2\space moles$ of $ZnO.$
So, $0.5128\space moles$ produce $0.5128\space moles$ of $ZnO.$
So, mass of $ZnO = (0.5128)\times 81 = 41\space g$
As percentage yield $=80\%$
$\Rightarrow$ Mass of $ZnO = \dfrac{80}{100} \times 41 = 33.4\space g$

$2Ag{ NO } _{ 3 }+Cu\rightarrow Cu{ \left( { NO } _{ 3 } \right)  } _{ 2 }+2Ag$
What is the percent yield when $0.17\ g$ of $Ag{NO} _{3}$ in aqueous solution reacts with excess copper to produce $0.08\ g$ $Ag$? (At. mass of $Ag=107\ g/mol$) 

  1. $74$%

  2. $47$%

  3. $89$%

  4. $65$%


Correct Option: A
Explanation:

$ 2AgNO _3 \space + \space Cu \rightarrow Cu(NO _3) _2 \space + \space 2Ag $

Percentage of Ag in $AgNO _3 = \dfrac{108 \times 100}{108 + 14 + 48} = \dfrac{108}{170} \times 100 = \dfrac{1080}{17} = 63.52\%$

So, amount of Ag produced $= \dfrac{63.52}{100} \times 0.17 = 0.108\space g$

$\%$ Yield $= \dfrac{0.08}{0.108} \times 100 = 74\%$

$Zn+{H} _{2}{SO} _{4}\rightarrow Zn{SO} _{4}+{H} _{2}$
A reaction of zinc metal with sulfuric acid produces $1.5\times {10}^{-2}\ mol$ of $Zn{SO} _{4}$ from $2.0\times {10}^{-2}\ mol$ of $Zn$.
What was the percent yield of this reaction?

  1. $25$%

  2. $75$%

  3. $33$%

  4. $67$%


Correct Option: B
Explanation:

$1\space mole$ of Zn react with $1\space mole$ of $H _2SO _4$ produce $1\space mole$ of $ZnSO _4$.

So, to produce $1.5 \times 10^{-2} \space ZnSO _4$, $\space 1.5 \times 10^{-2} \space moles$.of zinc is needed.
Here, Actual Yield $= 1.5 \times 10^{-2} \space moles$
Theoretical Yield $= 2 \times 10^{-2} \space moles$
$\Rightarrow $ Percent Yield $= \dfrac{1.5\times 10^{-2}}{2 \times 10^{-2}} \times 100 = 75\%$

$4Co+3{ O } _{ 2 }\rightarrow 2C{ o } _{ 2 }{ O } _{ 3 }$
$66.8\ g$ of Cobalt reacted with oxygen and $70.50\ g$ of $C{ o } _{ 2 }{ O } _{ 3 }$ was collected after the reaction was completed. Calculate the percent yield. (At. mass of $Co=59\ g/mol$)

  1. $75$%

  2. $80$%

  3. $85$%

  4. $90$%


Correct Option: A
Explanation:

$ 4Co \space + \space 3O _2 \rightarrow 2Co _2O _3$

$4\space moles$ of cobalt produce $2\space moles$ of $Co _2O _3$

$\Rightarrow 2\space mole$  $Co \rightarrow$  $1 \space mole \space Co _2O _3$ 

$\Rightarrow 2\times 59 \rightarrow (2\times 59 + 3\times 16)$

$\Rightarrow 118\space g \space Co \rightarrow 166\space g \space Co _2O _3$

$\Rightarrow 66.8\space g \space \rightarrow (x)$

$\Rightarrow x = \dfrac{166 \times 66.8}{118} = 93.97\space g$

$\%$ Yield $= \dfrac{70.50}{93.97}\times 100 \approx 75\%$


A reaction produced $30.0$ grams of carbon dioxide. If the theoretical (expected) yield was $45$ grams, what is the percentage yield?

  1. $15$%

  2. $30$%

  3. $67$%

  4. $150$%


Correct Option: C
Explanation:

Percentage yield of a compound is the ratio of actual yield to the excepted yield of the compound.

$\%$ Yield = $ \dfrac{\text{Actual Yield}}{\text{Expected Yield}} \times 100 = \dfrac{30}{45} \times 100 = \dfrac{2}{3}\times 100 = 67\%$

When the fertilizer plant completes its small batch process, they find that they have only collected $26\ mL$ of $NH _{3}$ which was supposed to be $100\ mL$. They are disappointed with this result due to the low yield. What is the % yield of their process?

  1. $26$%

  2. $38$%

  3. $52$%

  4. $76$%


Correct Option: A
Explanation:

$26$ ml of $NH _3$ yeild into 100 ml 

$=26/100$ 
$26%$

A reaction is known to have a percent yield of $80$%. It the actual yield is $25$ grams, what was the expected yield?

  1. $3.2\ grams$

  2. $19\ grams$

  3. $31.25\ grams$

  4. $190\ grams$


Correct Option: C
Explanation:

Percent yield is the ratio of actual yield to the expected yield of compound.

$\%$ Yield $= \dfrac{\text{Actual Yield}}{\text{Expected Yield}} \times 100$
$\Rightarrow 80 = \dfrac{25}{\text{Expected Yield}} \times 100$
$\Rightarrow$ Expected Yield $= \dfrac{25}{80} \times 100 = \dfrac{250}{8} = 31.259\space g$

A bacterium can convert xylose to lactic acid with a percent yield of $85$%. What mass of lactic acid would be produced from $50.0\ g$ sample of xylose by this bacterium? (At. mass of xylose$(C _5H _{10}O _5) =150\ g/mol,$ At. of lactic acid$(C _3H _6O _3)=90\ g/mol$).

  1. $58.8\ g$

  2. $85\ g$

  3. $50\ g$

  4. $42.5\ g$


Correct Option: D
Explanation:

Percent Yield is the ratio of actual yield of a substance to the theoretical yield.

Percent Yield $= \dfrac{\text{Actual Yield}}{\text{Theoretical Yield}} \times 100$
Given, Theoretical Yield $50\space g,$ Percent Yield $= 85$
So, $(\dfrac{85}{100} \times 50) =$ Actual Yield
$\Rightarrow$ Actual Yield $= 42.5 \space g$

A student conducts an experiment to produce a $Ca{CO} _{3}$ precipitate. The student collects $1.80\ g$ of product after predicting it should be possible to produce $2.00\ g$ of the product.
What is the student's percent yield for this experiment?

  1. $-10$%

  2. $+90$%

  3. $+10$%

  4. $+111$%


Correct Option: B
Explanation:

Percent Yield is the ratio of actual yield to the theoretical yield.

$\Rightarrow \% $ Yield $= \dfrac{E}{T} \times 100$
$\Rightarrow$ Here, $E = 1.8 \space g; \space T = 2\space g$
So, percent yield $= \dfrac{1.8}{2} \times 100 = 90\%$

If $100\ mL$ of the acid is neutralised by $100\ mL$ of $4\ M\ NaOH$, the purity of concentrated $HCl$ (sp. gravity $= 1.2)$ is:

  1. $12$%

  2. $98$%

  3. $73$%

  4. $43$%


Correct Option: A
Explanation:

No of equivalent of acid $=$ No of equivalent of base.

$100 \times m _1=100 \times4$
$m _1=4m$
$4$mole of $HCL$ in $1$liter of water..........$(1)$
$\frac{{\rho HCl}}{{\rho {H _2}O}} = k$
density of $HCl=1200gm/l$...........$(2)$
purity of $HCl\Rightarrow 1200gm\,\,in\,1liter$
purity of $HCl$ in gram in $1$ liter $\Rightarrow 1200gm$
purity of $HCl \Rightarrow$ $\frac{{4 \times 36.5}}{{1200}} \times 100 \Rightarrow  \sim 12\% $
hence, purity of $HCl$ is $ \sim 12\% $
so option $A$ is correct.

$50\ g$ of a sample of $NaOH$ required for complete neutralisation of $1\ litre\ N\ HCl$. What is the percentage purity of $NaOH$?

  1. $80$

  2. $70$

  3. $60$

  4. $50$


Correct Option: A
Explanation:

for complete neutralisation we have

moles of acid = moles of base
moles of acid = normality $$ volume = $1 *1$
moles of base = 1 =$\dfrac{given mass}{molecular mass}$

so, given mass = moles $$ molecular mass = 1$*$ = 40

percentage purity = $\dfrac{40 }{50} * 100$ = 80 %

In a reaction vessel,100 g $H _2$ and 100 g $Cl _2$ are inbred and suitable conditions are provided for take following reaction: 

$H _2 ( g)+ Cl _2(g)\rightarrow 2HCl(g)$


The amount of $HCI$ formed (at 90% yield) will be:

  1. 36.8 g

  2. 62.5 g

  3. 80 g

  4. 91.98 g


Correct Option: D
Explanation:

100 g of $H _2$ = 50 mole

100 g of $Cl _2$ = 1.4 mole
According to the reaction 1 mole of $H _2$ is reacting with 1  mole of $Cl _2$
So, $Cl _2$ is th limiting reagent 
So, moles of $HCl$ formed = 2 $\times$ 1.4 mole
 
Amount of $HCl$ = 2.8 $\times$ 36.5 = 102.2 g

Since, the reaction is giving 90 % yield
therefore , amount of $HCl$ formed = $102.2 \times \dfrac{90}{100}$

Amount of HCl formed = 91.98 g

Hence, the correct option is $(D)$.

In the Haber process:
$N _2(g) + 3H _2(g) \rightarrow 2NH _3(g) $
$30 L$ of $H _2$ and $30 L$ of ${N _2}{^-}$ were taken for reaction which yielded only $50\%$ of expected product. What will be the composition of the gaseous mixture in the end?

  1. 20 L $NH _3$, 25 L $N _2$ and 20 L $H _2$

  2. 10 L $NH _3$, 25 L $N _2$ and 15 L $H _2$

  3. 20 L $NH _3$, 10 L $N _2$ and 30 L $H _2$

  4. 20 L $NH _3$, 25 L $N _2$ and 15 L $H _2$


Correct Option: A

$0.5\ g$ of impure ammonium chloride was heated with caustic soda solution to evolve ammonia gas, the gas is absorbed in $150\ mL$ of $N/5\ H _{2}SO _{4}$ solution. Excess sulphuric acid required $20\ mL$ of $1\ N\ NaOH$ for complete neutralization. The percentage of $NH _{3}$ in the ammonium chloride is:

  1. $68$%

  2. $34$%

  3. $48$%

  4. $17$%


Correct Option: B
Explanation:

$(NH _3)Cl+NaOH\rightarrow +NH _3\uparrow +H _2O$

excess $H _2SO _4$ reg. $20ml$ of $1N\,\,NaOH$
$\Rightarrow $ moles of $H _2SO _4=(20\times 1)milimoles$
                                 $=20m\,mol$
added amount of $H _2SO _4=\dfrac{N}{5}\times 150ml=30mmol$
amount of $h _2so _4$ reacted with $nh _3=0.01MOLE$
$\Rightarrow 0.01\, mole$ of $NH _3$ present in $(NH _4)Cl$
So, $\%purity=\dfrac{(0.01)\times 17}{0.5}\times 100=34\%$

A sample of $CaCO _3$ is 50% pure. On heating $1.12 L$ of $CO _2$ (at STP) is obtained. Residue left (assuming non-volatile impurity) is

  1. 7.8 g

  2. 3.8 g

  3. 2.8 g

  4. 8.9 g


Correct Option: A
Explanation:

Solution:- (A) $7.8 \; g$

Volume of $C{O} _{2}$ formed $= 1.12 \; L$
At STP, volume of $1$ mole of gas $= 22.4 \; L$
$\therefore$ No. of moles of $C{O} _{2}$ formed $= \cfrac{1.12}{22.4} = 0.05 \text{ mol}$
Heating of $CaC{O} _{3}$-

$CaC{O} _{3} \longrightarrow CaO + C{O} _{2}$
From the above reaction,
$1$ mole of $C{O} _{2}$ is formed on heating $1$ mole of $CaC{O} _{3}$.
Therefore,
$0.05$ mole of $C{O} _{2}$ is formed on heating $0.05$ mole of $CaC{O} _{3}$.

Molecular weight of $CaC{O} _{3} = 100 \; g$
$\therefore$ Weight of $CaC{O} _{3} = 0.05 \times 100 = 5 \; g$
As the sample was $50 \%$ pure.
Thus the $50 \%$ of the sample was heated in the form of $CaC{O} _{3}$.
Amount of sample left unreacted $= 5 \; g$
Also,
No. of moles of $CaO$ formed $= 0.05$
Molecular weight of $CaO = 56 \; g$
Weight of $CaO = 56 \times 0.05 = 2.8 \; g$
Therefore,
Amount of residue left $= 5 + 2.8 = 7.8 \; g$

In ayurvedic preparation of swarnabhasma, what purity of golden will be used?

  1. $42$ % gold + silver

  2. $91$ % amalgam

  3. $24$ carat

  4. $58.5$% copper


Correct Option: A
Explanation:

Swarna Bhasma is prepared from Gold. It is used in Ayurvedic treatment of infertility, asthma, tissue wasting, poisoning etc. This medicine should only be taken strictly under medical supervision.

In the given question option A is the correct answer.

One mole of photons is known as one Einstein of radiation. According to Stark-Einstein law of photochemical equivalence, one mole of reactant absorbs one Einstien of energy. For a photochemical reaction, a term called quantum yield is defined as:
Quantum yield $ (\phi) = \dfrac {No. \,of \,moles \,of \,reactant \,converted} {No. \,of \,Einstein \,absorbed} $
The correct statement(s) is/are:

  1. for a chain reaction $\phi _{gas} >> \phi _{solution}$

  2. in a photochemical chain reaction $\phi >> 1$

  3. in a photochemical chain reaction $\phi << 1$

  4. for a chain reaction $\phi _{gas} << \phi _{solution}$


Correct Option: A,B
Explanation:
$Quantum \ Yield= \cfrac {Number \ of \ moles \ of \ reactant \ converted}{Number \ of \ einstein \ absorbed}$
An einstein of radiation $=$ one mole of photons
For a chain reaction, $\phi _{gas} >> \phi _{solution}$
and ln photochemical chain reaction $\phi >>1$

A sample of $CaC{O _3}$ is $50\% $ pure. On heating $1.12{\text{ }}L$ of $C{O _2}$ (at STP) is obtained. Residue left (assuming non-volatile impurity) is:

  1. 7.8 g

  2. 3.8 g

  3. 2.8 g

  4. 8.9 g


Correct Option: A
Explanation:

No. of moles of $CO _2$ evolved $=\cfrac{1.12}{22.4}=0.05$ $moles$

$CaCO _3(s)\overset { \Delta }{ \longrightarrow } CaO\downarrow+CO _2\uparrow$
                             $0.05$        $0.05$ $moles$
So, $0.05$ $moles$ of $CaCO _3$ have  reacted.
Mass $=0.05\times 100=5$ $gm=50\%$ of $CaCO _3$ sample
Total weight $=2\times 5$ $gm=10$ $gm$
Residue left by $CaCO _3=5$ $gm$
Residue left by $CaO=56\times 0.05=2.8$ $gm$
Toatl residue $=5+2.8=7.8$ $gm$

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