Understanding the Problem
The reaction provided is the thermal decomposition of Sodium Bicarbonate ($NaHCO_3$): $2NaHCO_3 \rightarrow Na_2CO_3 + H_2O + CO_2$
We are given an impure sample of $16.8\,g$ of $NaHCO_3$, which produces $9\,g$ of $Na_2CO_3$. We need to calculate the purity of the sample and the volume of $CO_2$ gas produced.
Atomic Masses
To solve this, we need the molar masses:
- $Na: 23, H: 1, C: 12, O: 16$
- Molar mass of $NaHCO_3 = 23 + 1 + 12 + (3 \times 16) = 84\,g/mol$
- Molar mass of $Na_2CO_3 = (2 \times 23) + 12 + (3 \times 16) = 106\,g/mol$
i. Calculating Percentage Purity
- Calculate moles of $Na_2CO_3$ produced: $n(Na_2CO_3) = \frac{mass}{molar mass} = \frac{9\,g}{106\,g/mol} \approx 0.0849\,mol$
- Use stoichiometry to find theoretical $NaHCO_3$: From the balanced equation, $1\,mol$ of $Na_2CO_3$ requires $2\,mol$ of $NaHCO_3$. $n(NaHCO_3) = 2 \times 0.0849 = 0.1698\,mol$
- Find mass of pure $NaHCO_3$: $mass = 0.1698\,mol \times 84\,g/mol \approx 14.26\,g$
- Percentage Purity: $Purity = \frac{Pure\,mass}{Given\,mass} \times 100 = \frac{14.26}{16.8} \times 100 \approx 84.88\%$
ii. Calculating Volume of $CO_2$
From the stoichiometry, $1\,mol$ of $Na_2CO_3$ produces $1\,mol$ of $CO_2$. Therefore, $n(CO_2) = 0.0849\,mol$.
Using Ideal Gas Law: $PV = nRT$
- $P = 1\,atm$
- $R = 0.0821\,L\cdot atm/(K\cdot mol)$
- $T = 27 + 273 = 300\,K$
- $V = \frac{nRT}{P} = \frac{0.0849 \times 0.0821 \times 300}{1} \approx 2.09\,L$