Oxygen gas, generated by the reaction 2KClO₃(s) → 2KCl(s) + 3O₂(g), is collected over water at 27°C in a 3.72-L vessel at a total pressure of 730. torr. (The vapor pressure of H₂O at 27°C is 26.0 torr.) How many moles of KClO₃ were consumed in the reaction?

Question

Quizplus · Accepted Answer

First, we need to calculate the partial pressure of oxygen gas in the container. We can use Dalton's law of partial pressures to do this:

P(O2) = P(total) - P(H2O)

P(O2) = 730. torr - 26.0 torr = 704. torr

Next, we need to use the ideal gas law to calculate the number of moles of oxygen gas in the container:

PV = nRT

n = PV/RT

n = (704. torr)(3.72 L)/(0.0821 L·atm/mol·K)(300 K)

n = 0.0933 mol

Finally, we can use stoichiometry to determine the number of moles of KClO3 consumed in the reaction:

2 mol KClO3 → 3 mol O2

x mol KClO3 → 0.0933 mol O2

x = (0.0933 mol O2)(2 mol KClO3/3 mol O2)

x = 0.0622 mol KClO3

Therefore, the answer is C) 0.0933 mol.

Oxygen Gas, Generated by the Reaction 2KClO3(s) → 2KCl(s)

Oxygen Gas, Generated by the Reaction 2KClO₃(s) → 2KCl(s)