Example Problems (Quantitative Chemistry)

1. An excess of $Zn$ is added to $100 \, cm^{3}$ of $1.0 \, mol/dm^{3}$ of hydrochloric acid. The equation for the reaction is: $$Zn \,\,\, + \,\,\, HCl \,\,\, \rightarrow \,\,\, ZnCl_{2} \,\,\, + \,\,\, H_{2} \uparrow$$ What is the maximum volume of hydrogen evolved at room temperature and pressure?
a) $1.2 \, dm^{3}$
b) $2.0 \, dm^{3}$
c) $2.4 \, dm^{3}$
d) $24 \, dm^{3}$

Observe that one $H_{2}$ molecule is evolved when one $Zn$ atom reacts with one $HCl$ molecule. So, the total number of $H_{2}$ molecules evolved must be the same as the molecules of $HCl$ or $Zn$ that participate in the reaction. Therefore, we will adopt the following strategy to solve this problem:
First, find out the number of moles of $H_{2}$ gas evolved by finding out how many $HCl$ or $Zn$ that participate in the reaction.
Second, we use the relationship $24 \, dm^{3} \, \equiv \, 1 \, mole$ for any gas at r.t.p. to convert moles of $H_{2}$ gas into volume.

Solution: Not enough information has been given to find moles of $Zn$ that participate in the reaction. So, we try to find the moles of $HCl$.
$$\begin{array}{l} \mbox{concentration of} \, HCl \, \mbox{solution} &=& \frac{\mbox{number of moles of} \, HCl \, \mbox{molecules}}{\mbox{volume of the solution}} \\ \mbox{number of moles of} \, HCl \, \mbox{molecules} &=& \mbox{concentration of} \, HCl \, \mbox{solution} \times \mbox{volume of the solution} \\ &=& 1.0 \, mol/dm^{3} \times 100 \, cm^{3} \\ &=& 1.0 \, mol/dm^{3} \times 100 \, \times 10^{-3} dm^{3} \\ &=& 0.1 \, mol \end{array}$$ From the reaction equation, it takes one $HCl$ molecule to produce one $H_{2}$ molecule; therefore, $0.1 \, mol$ of $HCl$ should produce $0.1 \, mol$ of $H_{2}$ gas.
Therefore, $0.1 \, mol$ of $H_{2}$ gas is produced in the reaction. Now, its time to use $24 \, dm^{3} \, \equiv \, 1 \, mole$ for a gas at r.t.p. $$\begin{array}{l} 1 \, mole \, of \, H_{2} \, gas\, & \equiv & 24 \, dm^{3} \\ 0.1 \, mole \, of \, H_{2} \, gas\, & \equiv & \frac{24 \, dm^{3}}{1 \, mole} \times 0.1 \, moles = 2.4 \, dm^{3} \end{array}$$ Hence, $H_{2}$ gas produced is $2.4 \, dm^{3}$

2. Neutralization reaction
Sodium hydroxide reacts with sulphuric acid according to the following equation: $$2NaOH \, (aq) \,\,\, + \,\,\, H_{2}SO_{4} \, (aq) \,\,\, \rightarrow \,\,\, Na_{2}SO_{4} \,\,\, + \,\,\, H_{2}O$$ Which volume of $0.4 \, mol/dm^{3}$ of sodium hydroxide reacts with $50 \, cm^{3}$ of $0.1 \, mol/dm^{3}$ sulphuric acid?
a) $12.5 \, cm^{3}$
b) $25.0 \, cm^{3}$
c) $50.0 \, cm^{3}$
d) $100.0 \, cm^{3}$

Solution: From the equation, it is clear that it takes two $NaOH$ molecules to react with one $H_{2}SO_{4}$ molecule. Here's how we solve this problem.
Step 1: we find the number of $H_{2}SO_{4}$ molecules in the solution;
Step 2: we find the number of $NaOH$ molecules in the solution (this will be twice the number of $H_{2}SO_{4}$ molecules);
Step 3: we find the volume of $NaOH$ needed using the given concentration and number of molecules of $NaOH$ found in step 2.
$$\begin{array}{l} \mbox{concentration of} \, H_{2}SO_{4} \, \mbox{solution} &=& \frac{\mbox{number of moles of} \, H_{2}SO_{4} \mbox{molecules}}{\mbox{volume of} \, H_{2}SO_{4}} \\ \mbox{number of moles of} \, HCl \, \mbox{molecules} &=& \mbox{concentration of} \, H_{2}SO_{4} \, \mbox{solution} \times \mbox{volume of the solution} \\ &=& 0.1 \, mol/dm^{3} \times 50 \, cm^{3} \\ &=& 0.1 \, mol/dm^{3} \times 50 \, \times 10^{-3} dm^{3} \\ &=& 0.005 \, mol \end{array}$$ It will take $2 \times 0.005 \, moles\ = 0.01 \, moles$ of $NaOH$ to react completely with the given $H_{2}SO_{4}$. We now need to find volume of $NaOH$ using the given concentration.
$$\begin{array}{l} \mbox{concentration of} \, NaOH \, \mbox{solution} &=& \frac{\mbox{number of moles of} \, NaOH \mbox{molecules}}{\mbox{volume of the} \, NaOH \, \mbox{solution}} \\ \Rightarrow \mbox{volume of the} \, NaOH \, \mbox{solution} &=& \frac{\mbox{number of moles of} \, NaOH}{\mbox{concentration of} \, NaOH \, \mbox{solution}} \\ &=& \frac{0.01 \, mol}{0.4 \, mol/dm^{3}} \\ &=& 0.025 \, dm^{3} \\ &=& 25 \, cm^{3} \end{array}$$ Hence, $25 \, cm^{3}$ of $NaOH$ of the given concentration reacts with the given sulphuric acid.

3. Sodium hydrogencarbonate undergoes thermal decomposition according to the following equation: $$2NaHCO_{3} \,\,\, \rightarrow \,\,\, Na_{2}CO_{3} \,\,\, + \,\,\, CO_{2} \,\,\, + \,\,\, H_{2}O$$ What is the maximum mass of sodium carbonate that can be made from 0.100 moles of sodium hydrogencarbonate?

Solution: From the equation, it is clear that it takes two $2NaHCO_{3}$ molecules to produce one $Na_{2}CO_{3}$ molecule. Therefore, $0.100$ moles of $2NaHCO_{3}$ will produce $0.100/2 \, moles = 0.05 \, moles$ of $Na_{2}CO_{3}$.

Molecular mass of $Na_{2}CO_{3}$ = $(2 \times 23)+(12)+(3 \times 16) \, amu$ = $106 \, amu$
So, $1 \, mole$ of $Na_{2}CO_{3}$ amounts to $106 \, gm$
$\Rightarrow 0.05 \, moles$ of $Na_{2}CO_{3}$ amounts to $0.05 \times 106 \, gm = 5.30 \, gm$

$5.30 \, gm$ of $Na_{2}CO_{3}$ can be produced using $0.100 \, moles$ of $NaHCO_{3}$.