Chemistry

Mole Concept and Stoichiometry

365 Questions

The mole concept is a core area of chemistry focusing on the quantitative relationships between reactants and products in chemical reactions. Questions cover molecular mass calculations, stoichiometric conversions, and determining the number of molecules. This topic is essential for most science entrance and competitive examinations.

molecular mass calculationsstoichiometric conversionsmole fraction problemsheavy water reactionsoxidation calculations

Mole Concept and Stoichiometry Questions

Multiple choice chemistry introduction to analytical chemistry interpreting a balanced chemical reaction percentage yield stoichiometric calculations

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%
Reveal answer Fill a bubble to check yourself
D Correct answer
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 $MgC{O} _{3} = 84.3 \% \approx 84 \%$

Multiple choice chemistry introduction to analytical chemistry interpreting a balanced chemical reaction percentage yield stoichiometric calculations

$10g$ 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. $80%$
Reveal answer Fill a bubble to check yourself
B Correct answer
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= $\cfrac {4.2}{5.6}\times 100= 75$%

Multiple choice chemistry introduction to analytical chemistry interpreting a balanced chemical reaction percentage yield stoichiometric calculations

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. $60%$
Reveal answer Fill a bubble to check yourself
D Correct answer
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 $60$%.
Multiple choice chemistry introduction to analytical chemistry interpreting a balanced chemical reaction percentage yield stoichiometric calculations

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
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

The question has garbled chemical formulas (FeSO4a, SO4 instead of S, O4). Assuming it means Fe2(SO4)3 and FeSO4: Molar mass of Fe2(SO4)3 = 400 g/mol, FeSO4 = 152 g/mol. For equal moles (let n = 1), mass of Fe2(SO4)3 = 400 g, mass of FeSO4 = 152 g. Total pure = 552 g. % of Fe2(SO4)3 in pure = (400/552) × 100 = 72.5%. In total mixture (600 g): 400/600 = 66.7%. The answer depends on whether we ask % in pure mixture or total sample.

Multiple choice chemistry introduction to analytical chemistry interpreting a balanced chemical reaction percentage yield stoichiometric calculations

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. $4.5*{10^{20}}$
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

Assuming 100% purity, 1 g gold = 1/197 mol. Atoms = (1/197) * 6.022 * 10^23 = 3.05 * 10^21. The provided answer 2.6 * 10^21 implies a purity of ~85%.

Multiple choice chemistry introduction to analytical chemistry interpreting a balanced chemical reaction percentage yield stoichiometric calculations

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%
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

CaO (56 g/mol): 2.46 g = 0.0439 mol. CaCl2 (111 g/mol): 3.7 g = 0.0333 mol. Theoretical yield = 0.0439 * 111 = 4.87 g. % Yield = (3.7 / 4.87) * 100 = 76%.

Multiple choice chemistry introduction to analytical chemistry interpreting a balanced chemical reaction percentage yield stoichiometric calculations

$4$ 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. $100\%$
Reveal answer Fill a bubble to check yourself
B Correct answer
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= $\cfrac {49}{98}\times 100=50$% .

Multiple choice chemistry introduction to analytical chemistry interpreting a balanced chemical reaction percentage yield stoichiometric calculations

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. $50$%
Reveal answer Fill a bubble to check yourself
D Correct answer
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 $\dfrac{0.333}{0.667} \times 100 = 50\%$

Multiple choice chemistry introduction to analytical chemistry interpreting a balanced chemical reaction percentage yield stoichiometric calculations

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. $90\%$
Reveal answer Fill a bubble to check yourself
A Correct answer
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 $= 92.5\%$

Multiple choice chemistry introduction to analytical chemistry interpreting a balanced chemical reaction percentage yield stoichiometric calculations

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
Reveal answer Fill a bubble to check yourself
D Correct answer
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

Multiple choice chemistry introduction to analytical chemistry interpreting a balanced chemical reaction percentage yield stoichiometric calculations

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
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

MgCO3 -> MgO + CO2. 8 g MgO = 8/40 = 0.2 mol. This requires 0.2 mol MgCO3 = 0.2 * 84 = 16.8 g. Purity = (16.8 / 20) * 100 = 84%.

Multiple choice chemistry introduction to analytical chemistry interpreting a balanced chemical reaction percentage yield stoichiometric calculations

$^{14} _6C\rightarrow ^{14} _7N+X$
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
Reveal answer Fill a bubble to check yourself
C Correct answer
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

Multiple choice chemistry introduction to analytical chemistry interpreting a balanced chemical reaction percentage yield stoichiometric calculations

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. $3240\ g$
Reveal answer Fill a bubble to check yourself
A Correct answer
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 $ZnO = \dfrac{80}{100} \times 41 = 33.4\space g$

Multiple choice chemistry introduction to analytical chemistry interpreting a balanced chemical reaction percentage yield stoichiometric calculations

$2Ag{ NO } _{ 3 }+Cu\rightarrow Cu{ \left( { NO } _{ 3 } \right)  } _{ 2 }+2Ag$
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. $65$%
Reveal answer Fill a bubble to check yourself
A Correct answer
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 $= \dfrac{0.08}{0.108} \times 100 = 74\%$

Multiple choice chemistry introduction to analytical chemistry interpreting a balanced chemical reaction percentage yield stoichiometric calculations

$Zn+{H} _{2}{SO} _{4}\rightarrow Zn{SO} _{4}+{H} _{2}$
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. $67$%
Reveal answer Fill a bubble to check yourself
B Correct answer
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 $= \dfrac{1.5\times 10^{-2}}{2 \times 10^{-2}} \times 100 = 75\%$