Question: Calculate the volume of gas that would be collected if the student heated $1.312$ g of $Ca(NO_3)_2$ at $25.0$ °C and $100$ kPa. Give your answer in $dm^3$ to $3$ significant figures.
Mass: $Ca = 40$; $N = 14$; $O = 16$
$R = 8.31$ J K$^{-1}$ mol$^{-1}$
1 Recall the equation
$2 Ca(NO_3)_2 \rightarrow 2 CaO(s) + 4 NO_2(g) + O_2$
1. **State the problem:** Calculate the volume of gas collected when $1.312$ g of $Ca(NO_3)_2$ decomposes at $25.0$ °C and $100$ kPa.
2. **Recall the balanced chemical equation:**
$$2 Ca(NO_3)_2 \rightarrow 2 CaO(s) + 4 NO_2(g) + O_2(g)$$
3. **Calculate the molar mass of $Ca(NO_3)_2$: **
$$M = 40 + 2(14 + 3 \times 16) = 40 + 2(14 + 48) = 40 + 2 \times 62 = 40 + 124 = 164 \text{ g/mol}$$
4. **Calculate moles of $Ca(NO_3)_2$ decomposed:**
$$n = \frac{1.312}{164} = 0.008 \text{ mol}$$
5. **Calculate moles of gas produced:**
From the equation, $2$ moles of $Ca(NO_3)_2$ produce $4 + 1 = 5$ moles of gas ($4$ moles $NO_2$ and $1$ mole $O_2$).
So, moles of gas produced per mole of $Ca(NO_3)_2$:
$$\frac{5}{2} = 2.5$$
Therefore, moles of gas from $0.008$ mol $Ca(NO_3)_2$:
$$0.008 \times 2.5 = 0.02 \text{ mol}$$
6. **Use ideal gas law to find volume:**
$$PV = nRT$$
Rearranged:
$$V = \frac{nRT}{P}$$
Convert temperature to Kelvin:
$$T = 25.0 + 273.15 = 298.15 \text{ K}$$
Convert pressure to Pa:
$$P = 100 \text{ kPa} = 100000 \text{ Pa}$$
Calculate volume in $m^3$:
$$V = \frac{0.02 \times 8.31 \times 298.15}{100000} = \frac{49.5}{100000} = 0.000495 \text{ m}^3$$
Convert to $dm^3$ (1 $m^3 = 1000$ $dm^3$):
$$V = 0.000495 \times 1000 = 0.495 \text{ dm}^3$$
7. **Final answer:**
$$\boxed{0.495 \text{ dm}^3}$$
This is the volume of gas collected to 3 significant figures.