Question

Oxygen gas, generated by the reaction 2KClO3 (s) 2KCl(s) 3O2 (g) is collected over water at 27°C in a 1.35-L vessel at a total pressure of 1.00 atm. (The vapor pressure of H2O at 27°C is 26.0 torr.) How many moles of KClO3 were consumed in the reaction?

Answer 1

Answer:

$n_{KClO_3}=0.0353molKClO_3$

Explanation:

Hello!

In this case, according to the experiment, we first realize that the pressure of oxygen is:

$p_{O_2}=1.00atm-26.0/760 atm=0.966atm$

Next, we compute the moles based off the ideal gas law:

$PV=nRT\\\\n=\frac{PV}{RT}=\frac{0.966atm*1.35L}{0.08206\frac{atm*L}{mol*K}*300K}=0.0530molO_2$

Thus, since there is 3:2 mole ratio between oxygen and potassium chlorate, we can compute the consumed moles of the latter as shown below:

$n_{KClO_3}=0.0530molO_2*\frac{2molKClO_3}{3molO_2} \\\\n_{KClO_3}=0.0353molKClO_3$

Best regards!