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

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. $150$%
Reveal answer Fill a bubble to check yourself
C Correct answer
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\%$

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

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. $+111$%
Reveal answer Fill a bubble to check yourself
B Correct answer
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 $= \dfrac{1.8}{2} \times 100 = 90\%$

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

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

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 $H _2$
Reveal answer Fill a bubble to check yourself
A Correct answer
Multiple choice chemistry introduction to analytical chemistry interpreting a balanced chemical reaction percentage yield stoichiometric calculations

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
Reveal answer Fill a bubble to check yourself
A Correct answer
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 $= 5 + 2.8 = 7.8 \; g$

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

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
Reveal answer Fill a bubble to check yourself
A Correct answer
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 $=5+2.8=7.8$ $gm$

Multiple choice chemistry states of matter: gaseous and liquid states gay lussac's law gas laws states of matter

In the reaction ${ N } _{ 2 }+{ 3H } _{ 2 }\longrightarrow { 2H } _{ 3 }$, ratio by volume of ${ N } _{ 2 },{ H } _{ 2 }$ and $ { NH } _{ 3 }$ is $1:3:2$. This illustrates law of :

  1. definite proportions

  2. multiple proportions

  3. reciprocal proportions

  4. gaseous proportions
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation

As per the Gay Lussac's law of combining volume of gases, the volumes of gaseous reactants and gaseous products bear a simple whole number ratio with each other if they are measured at same temperature and pressure.
In the reaction ${ N } _{ 2 }+{ 3H } _{ 2 }\longrightarrow { 2H } _{ 3 }$, ratio by volume of $N _2,  H _2$ and $ NH _3$ is $1:3:2$. This illustrates law of Gay Lussac's law of combining volumes of gases.

Multiple choice chemistry states of matter: gaseous and liquid states gay lussac's law gas laws states of matter

In the reaction $N _{2}+3H _{2}\rightarrow 2NH _{3} $, the ratio by volume of $N _{2},\ H _{2} :$ and$: NH _{3}$ is $1 : 3 : 2$. 


This illustrates the law of:

  1. definite proportion

  2. multiple proportion

  3. reciprocal proportion

  4. gaseous volumes
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation

In the reaction, $N _{2}+3H _{2}\rightarrow 2NH _{3} $, the ratio by volume of $N _{2},\ H _{2} $ and $ NH _{3}$ is $1 : 3 : 2$. This illustrates the law of Gaseous volumes or Gay Lussac's law of combining volumes of gases.

According to this law, when gases react together to produce gaseous products, the volume of reactants and products bear a simple whole-number ratio with each other, provided volumes are measured at the same temperature and pressure.

So, the correct option is $D$.

Multiple choice chemistry further aspects of equilibria indicators and acid-base titration study of indicators properties of acids and bases

In the mixture of $NaHCO _{4}$ and $Na _{2}CO _{3}$, volume of a given $HCl$ required is $x\ mL$ with phenolphthalein indicator and $y\ mL$ with methyl orange indicator in same titration. Hence, volume of $HCl$ for complete reaction of $Na _{2}CO _{3}$ present in the original mixture is

  1. $2x$
  2. $y$
  3. $x/2$
  4. $(y - x)$
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

With phenolphthalein, Na2CO3 is converted to NaHCO3 (half neutralization). With methyl orange, the NaHCO3 is converted to NaCl. The volume for the second half is equal to the first, so the total volume for Na2CO3 is 2x.

Multiple choice chemistry further aspects of equilibria indicators and acid-base titration study of indicators properties of acids and bases

$40\ mL$ of $0.05\ M\ Na {2}CO _{3}\cdot NaHCO _{3} \cdot 2H _{2}O$ (sesquicarbonate) is titrated against $0.05\ M\ HCl.\ x\ mL$ of $HCl$ is used when phenolphthalein is the indicator and $y\ mL\ HCl$ is used when methyl orange is the indicator in two separate titrations, hence $(y - x)$ is_______.

  1. $80\ mL$
  2. $30\ mL$
  3. $120\ mL$

  4. none of the above
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation
Titration of ${ Na } _{ 2 }{ CO } _{ 3 }.{ NaHCO } _{ 3 }.2{ H } _{ 2 }O$ with $HCl$ involves following reactions :
a) ${ Na } _{ 2 }{ CO } _{ 3 }+HCl\rightleftharpoons { NaHCO } _{ 3 }+NaCl$
b) ${ NaHCO } _{ 3 }+HCl\rightleftharpoons NaCl+{ H } _{ 2 }O+{ CO } _{ 2 }$
In step $a$, $40$ ml of $0.05M$ $HCl$ will react with $40$ ml of $0.05M$ ${ Na } _{ 2 }{ CO } _{ 3 }$ to form ${ NaHCO } _{ 3 }$ using phenolpthalein.
$\therefore$   $x=40$ ml
Now, in a separate titration $40$ ml of $0.05M$ $HCl$ will need to react with $0.05M$ ${ Na } _{ 2 }{ CO } _{ 3 }$ to form ${ NaHCO } _{ 3 }$. Now in the second step $b$ total $80$ ml of $0.05M$ $HCl$ will need to neutralise ${ NaHCO } _{ 3 }$ completely.
$\therefore$   $y=40+40\times 2=120$
$\therefore$   $y-x=120-40=80$ ml
Answer will be $A$.
Multiple choice chemistry study of compounds a. hydrogen chloride laboratory method of preparation of hydrochloric acid physical and chemical properties and uses of hydrochloric acid hydrogen chloride: occurrence, preparation, purification and identification

In which reaction equilibrium moves in left hand side when pressure is increased?

  1. $H _{2(g)} + Cl _{2(g)} \rightleftharpoons 2HCl _{(g)}$
  2. $2Mg _{(s)} + O _{2(g)} \rightleftharpoons 2MgO _{(s)}$
  3. $2H _{2}O _{(g)} \rightleftharpoons 2H _{2(g)} + O _{2(g)}$
  4. $N _{2(g)} + 3H _{2(g)} \rightleftharpoons 2NH _{3(g)}$
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

According to Le Chatelier's principle, increasing pressure shifts the equilibrium toward the side with fewer moles of gas. In reaction C, there are 2 moles of gas on the left and 3 moles on the right. Increasing pressure shifts it to the left (fewer moles).

Multiple choice chemistry chemical equilibrium equilibrium in chemical processes introduction to equilibrium chemical equilibrium and acids-bases

$4g \,H _2$ and $127g \,I _2$  are mixed and heated lit closed vessels until equilibrium is reached. If the equilibrium concentration of $HI$ is $0.05 \,M$ total number of moles present at equilibrium is:

  1. $3.25$
  2. $1.75$
  3. $2.25$
  4. $2.5$
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation

Given $H _2=4g$ & $I _2=127 g$

$\Rightarrow H _2=2$ mole & $I _2=0.5$ mole
$[HI] _{eqm}=0.05M$   $\therefore$ Moles of $HI=0.05$ mole
            $H _2\quad +\quad I _2\quad \rightleftharpoons\quad  2HI$
              $2$              $0.5$                  $0$         Initial
  $2-\cfrac {0.05}{2}$     $0.5-\cfrac {0.05}{2}$       $0.05$       Eqm
$\therefore$ Total moles at eqm,
$=\left(2-\cfrac {0.05}{2}\right)+\left(0.5-\cfrac {0.05}{2}\right)+0.05$
$=2.5$

Multiple choice chemistry chemical equilibrium equilibrium in chemical processes introduction to equilibrium chemical equilibrium and acids-bases

For the reaction ${ CO(g)+H } _{ 2 }O(g)\rightleftharpoons { CO } _{ 2 }(g)+{ H } _{ 2 }(g)$ at a given temperature the equilibrium amount of ${ CO } _{ 2 }(g)$ can be increased by:

  1. Adding a suitable catalyst

  2. Adding an inert gas

  3. Decreasing the volume of container
  4. Increasing the amount of $CO(g)$
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation

$CO _(g)^+\ H _2(g)\rightleftharpoons CO _{2(g)}+H _{2(g)}$

$\Delta x=0$   $\therefore$ Adding inert gas & decreasing volume will have no effect. by increasing amount of CO, we shift reaction forward and to more $CO _2$.

Multiple choice chemistry chemical equilibrium equilibrium in chemical processes introduction to equilibrium chemical equilibrium and acids-bases

The reactions $PCl 5 (g)  \rightleftharpoons  PCl _3(g) + Cl _2 (g) $ and $COCl _2 (g)  \rightleftharpoons  CO(g) + Cl _2(g)$ are simultaneously in equilibrium in an equilibrium box at constant volume. A few moles of CO(g) are later introduced into the vessel. After some time, the new equilibrium concentration of_______.

  1. PCl$ _5$ will remain unchanged
  2. Cl$ _2$ will be greater
  3. PCl$ _5$ will become less
  4. PCl$ _5$ will become greater
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

If CO is added 2$^{nd}$ equilibrium will proceed in the backward direction and concentration of Cl$ _2$ will decrease. This Cl$ _2$ will be further formed by the decomposition of PCl$ _5$.