Understanding Stoichiometry and Limiting Reagents
In chemical reactions, a limiting reagent is the reactant that is completely consumed first, thereby determining the maximum amount of product that can be formed. To solve this problem, we follow a systematic approach.
The Chemical Equation
First, we write the balanced chemical equation for the reaction between sodium carbonate ($Na_2CO_3$) and hydrochloric acid ($HCl$):
$$Na_2CO_3 + 2HCl \rightarrow 2NaCl + H_2O + CO_2$$
Step 1: Calculate Initial Moles
Molar Mass of $Na_2CO_3$: $(2 \times 23) + 12 + (3 \times 16) = 106 \text{ g/mol}$
Molar Mass of $HCl$: $1 + 35.5 = 36.5 \text{ g/mol}$
Moles of $Na_2CO_3$: $n = \frac{10.6 \text{ g}}{106 \text{ g/mol}} = 0.1 \text{ mol}$
Moles of $HCl$: $n = \frac{7.9 \text{ g}}{36.5 \text{ g/mol}} \approx 0.216 \text{ mol}$
(a) Identifying the Limiting Reagent
According to the balanced equation, 1 mole of $Na_2CO_3$ requires 2 moles of $HCl$. Therefore, 0.1 mole of $Na_2CO_3$ requires $0.1 \times 2 = 0.2$ moles of $HCl$.
Since we have 0.216 moles of $HCl$ (which is greater than 0.2), $Na_2CO_3$ is the limiting reagent.
- Unreacted $HCl$: $0.216 - 0.2 = 0.016 \text{ moles}$.
(b) Volume of $CO_2$ at NTP
From the equation, 1 mole of $Na_2CO_3$ produces 1 mole of $CO_2$.
- Moles of $CO_2 = 0.1 \text{ mol}$.
- At NTP (Normal Temperature and Pressure, 22.4 L/mol): $V = 0.1 \times 22.4 = 2.24 \text{ Liters}$.
(c) Mass of $NaCl$ Formed
From the equation, 1 mole of $Na_2CO_3$ produces 2 moles of $NaCl$.
- Moles of $NaCl = 0.1 \times 2 = 0.2 \text{ mol}$.
- Mass of $NaCl = 0.2 \text{ mol} \times 58.5 \text{ g/mol} = 11.7 \text{ g}$.