Question

KClO3 - 2KCl + 3O2 the O2 produced was collected by the displacement of water at 22oC at a total pressure of 760 torr. The volume of the gas collected was 1.20 liters, and the vapor pressure of water at 22 degree Celsius is 21 torr. Calculate the partial pressure of O2 in the gas collected and the mass of KClO3 in the sample that was composed

Answer 1

$$2 \mathrm{KClO_3} \rightarrow 2 \mathrm{KCl} + 3 \mathrm{O_2}$$ $$\mathrm{P_{total}} = \mathrm{P_{O_2}} + \mathrm{P_{H_2O}} = \mathrm{P_{O_2}} + 21 \text{ torr} = 760 \text{ torr}$$ $$\mathrm{P_{O_2}} = 739 \text{ torr} \cdot \frac{1 \text{ atm}}{760 \text{ torr}} = 0.972 \text{ atm}$$ $$\mathrm{n_{O_2}} = \frac{0.972 \text{ atm} \cdot 1.20 \text{ L}}{\frac{0.08206 \text{ L} \cdot \text{atm}}{\text{mol} \cdot \text{K}} \cdot 295 \text{ K}} = 4.82 \cdot 10^{-2} \text{ mol } \mathrm{O_2}$$ 4.82 \cdot 10^{-2} \text{ mol } \mathrm{O_2} \cdot \frac{2 \text{ mol KClO_3}}{3 \text{ mol } \mathrm{O_2}} = 3.21 \cdot 10^{-2} \text{ mol KClO_3} 3.21 \cdot 10^{-2} \text{ mol KClO_3} \cdot \frac{122.6 \text{ g KClO_3}}{1 \text{ mol KClO_3}} = 3.94 \text{ g KClO_3}

Answer provided by AssignmentExpert.com