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 \(NaHCO_{3}\).