Chemistry

Chemical Equilibrium and Stoichiometry

135 Questions

Chemical equilibrium and stoichiometry questions cover equilibrium constants, percent yield, and the Haber process. These concepts test your ability to calculate reaction outcomes and purity. Practice these to excel in physical chemistry sections of competitive tests.

Equilibrium constant calculationsStoichiometric yieldHaber processPercent purityHomogenous equilibrium

Chemical Equilibrium and Stoichiometry Questions

Multiple choice chemistry chemical equilibrium relationship between equilibrium constant, reaction quotient and gibbs energy law of mass action chemical equilibrium and acids-bases

Hydrolysis of sucrose gives glucose and fructose. The reaction takes place as: Sucrose $+ H _{2}O \rightleftharpoons$ Glucose $+$ Fructose. The equilibrium constant $K _{c}$ for this reaction is $2\times 10^{13}$ at $300\ K$. The $ \Delta G^{\circ}$ at $300\ K$ is:

  1. $7.64\times 10^{4}\ J\ mol^{-1}$
  2. $7.64\times 10^{-4}\ J\ mol^{-1}$
  3. $-7.64\times 10^{-4}\ J\ mol^{-1}$
  4. $-7.64\times 10^{4}\ J\ mol^{-1}$
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation

The relationship between the standard free energy change $(\Delta G^o)$ and the equilibrium constant $(K _p)$ is $(\Delta G^o)=-RTlnK _c$, where, $R$ is the ideal gas constant and $T$ is the temperature.


Given, $K _c=2\times 10^{13}, T=300K, R=8.314 :Jmol^{-1}K^{-1}$

Substituting these values in the above expression, we get

$(\Delta G^o)=-RTlnK _c=-8.314 \times 300 \times ln(2\times 10^{13})=-7.64\times 10^{4} J mol^{-1}$ 

Hence, the standard free energy change $(\Delta G^o)=-7.64\times 10^{4} J mol^{-1}$

Multiple choice chemistry chemical equilibrium relationship between equilibrium constant, reaction quotient and gibbs energy law of mass action chemical equilibrium and acids-bases

 $SO _2(g) + 1/2O _2 (g)\rightleftharpoons SO _3(g) \Delta H^o _{298} = 98.32 kJ/mole, \Delta S^o _{298} = 95.0 J/K/mole$.


 Find the $K _p$ for this above reaction at 298K:

  1. $K _P = 9.31 \times 10^{-12} atm^{1/2}$
  2. $K _P = 5.34 \times 10^{-13} atm^{1/2}$
  3. $K _P = 3.7 \times 10^{-13} atm^{1/2}$
  4. $K _P = 3.7 \times 10^{-14} atm^{1/2}$
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation
$\displaystyle \Delta G^0 = \Delta H^0 - T\Delta S^0  $
$\displaystyle  \Delta G^0 =  98.32 \times 1000 - 298 \times 95.0 = 70010 J/mol$
$\displaystyle  \Delta G^0 = -RTlnK _P$
$\displaystyle 70010 = - 8.314 \times 298 \times ln K $
$\displaystyle  ln K = -28.26$
$\displaystyle  K = 5.34 \times 10^{-13}$
Multiple choice introduction to mole gas laws and mole concept atoms and molecules chemistry mole concept chemical formula and mole concept

If of conservation of mass was to hold true, then 20.8 g of ${ BaCl } _{ 2 }$ on reaction with 9.8 g of ${ H } _{ 2 }{ SO } _{ 4 }$ will produce 7.3 g of $HCl$ and ${ BaSO } _{ 4 }$ equal to :

  1. 11.65 g

  2. 23.3 g

  3. 25.5 g

  4. 30.6 g
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

$BaCl _2+H _2SO _4 \longrightarrow BaSO _4+2HCl$


$1$ mole of $BaCl _2$ reacts with $1$ mole of $H _2SO _4$ to give $1$ mole of $BaSO _4$ and $2$ moles of $HCl$.


Here, moles of $BaCl _2=\dfrac{20.8}{208}=$ moles of $H _2SO _4= \dfrac{9.8}{98}=0.1$

$\therefore$ Moles of $BaSO _4$ formed $=0.1$

$\therefore$ Mass of $BaSO _4$ formed $=0.1 \times 233= 23.3 g$


i.e. $20.8+9.8=7.3+23.3=30.6$

Hence the correct option is B.

Multiple choice introduction to mole gas laws and mole concept atoms and molecules chemistry mole concept chemical formula and mole concept

Two acids $ H _{2}SO _{4} $ and $ H _{3}PO _{4} $ are neutralized separately by the same amount of an alkali when sulphate and dihydrogen orthophosphate are formed, respectively. Find the ratio of the masses of $ H _{2}SO _{4} $ and $ H _{3}PO _{4} $ 

  1. $ 1:1 $
  2. $ 1:2 $
  3. $ 2:1 $
  4. $ 2:3 $
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

$H _{2}SO _{4}+2NaOH\rightarrow Na _{2}SO _{4}+2H _{2}O$

$H _{3}PO _{4}+NaOH\rightarrow NaH _{2}PO _{4}+H _{2}O$

Equivalent of alkali $= 19$ eq of $H _{2}SO _{4}= 1g$ eq of $H _{3}PO _{4}$

Two acids must be reacting in the ratio of their equivalent masses

Eq. wt. of $H _{2}SO _{4}=\dfrac{98}{2}=49$

Eq. wt. of $H _{3}PO _{4}= \dfrac{98}{1}=98$

$\therefore $ ratio of masses of $H _{2}SO _{4}$ & $H _{3}PO _{4}$
$49:98=1:2$

$\Rightarrow 1:2$

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

$1.25$ 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. $80\%$
Reveal answer Fill a bubble to check yourself
D Correct answer
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 =$\cfrac{1}{1.25}\times 100$
$=80\%$

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

$12.5$ 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. $85$%
Reveal answer Fill a bubble to check yourself
C Correct answer
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 $C$ $(80$%$)$.

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

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

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

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

$^{14} _6C\rightarrow ^{14} _7N+X$
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

$Zn+{H} _{2}{SO} _{4}\rightarrow Zn{SO} _{4}+{H} _{2}$
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 $= \dfrac{1.5\times 10^{-2}}{2 \times 10^{-2}} \times 100 = 75\%$

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

$4Co+3{ O } _{ 2 }\rightarrow 2C{ o } _{ 2 }{ O } _{ 3 }$
$66.8\ g$ 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. $90$%
Reveal answer Fill a bubble to check yourself
A Correct answer
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 $= \dfrac{70.50}{93.97}\times 100 \approx 75\%$