Question

Given the following reaction: 2 KClO3(s) 2KClO3(s) + 3 O2(g)
What mass of KClO3 will produce 250mL of O2 gas, collected by water displacement, at an atmospheric pressure of 745 torr and a temperature of 30.0 C? The vapor pressure of water at 30.0 C is 31.8 torr.

Answer 1

Given the following reaction: $2\mathrm{KClO}_3(\mathrm{s}) \rightarrow 2\mathrm{KCl}(\mathrm{s}) + 3\mathrm{O}_2(\mathrm{g})$

What mass of $\mathrm{KClO}_3$ will produce $250~\mathrm{mL}$ of $\mathrm{O}_2$ gas, collected by water displacement, at an atmospheric pressure of 745 torr and a temperature of $30^{\circ}\mathrm{C}$ ? The vapor pressure of water at $30^{\circ}\mathrm{C}$ is 31.8 torr.

Solution: We must calculate the volume of gas at STP. We assume that gas can be described by Dalton's law, and then pressure of gas is equal to difference between atmospheric pressure and vapor pressure of water: $P = P_{B} - p_{w} = 745 - 31.8 = 713.2$ torr. According to the combined gas law, $\frac{P_0 \cdot V_0}{T_0} = \frac{P \cdot V}{T}$ where

$P_{0} = 760$ torr, $T_{0} = 273\mathrm{K}$ . Then, $V_{0} = V\cdot \frac{P\cdot T_{0}}{P_{0}\cdot T} = 250\cdot \frac{713.2\cdot 273}{760\cdot(30 + 273)} = 211.4\mathrm{mL} = 0.2114\mathrm{L}$

Molar mass of $\mathrm{KClO}_3$ is $M(KClO_3) = 39 + 35.5 + 16\cdot 3 = 122.5\mathrm{g / mol}$ .

According to the reaction equation we can make a proportion, where 22,4 L/mol is the STP molar volume:

$\mathrm{KClO}_3$ - $\mathrm{O}_2$

2·122.5 g/mol 3·22.4 L/mol

x grams 0.2114 L

Then, $m(KClO_3) = x = \frac{2 \cdot 122.5 \cdot 0.2114}{3 \cdot 22.4} = 0.771$ g;

Answer: $0.771\mathrm{g}$