Oxygen gas, generated by the reaction 2KClO₃(s) $\rightarrow$ 2KCl(s) + 3O₂(g), is collected over water at 27°C in a 3.78-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

To solve the problem, we first need to find the partial pressure of oxygen gas using Dalton's Law:

P(O2) = P(total) - P(H2O)
P(O2) = 730. torr - 26.0 torr
P(O2) = 704. torr

Next, we can use the ideal gas law to find the number of moles of oxygen gas:

PV = nRT
n = PV/RT
n = (704. torr * 3.78 L) / (0.0821 L•atm/mol•K * 300 K)
n = 0.0948 mol

According to the balanced chemical equation, 2 moles of KClO3 produce 3 moles of O2 gas. Therefore, the number of moles of KClO3 consumed in the reaction is:

n(KClO3) = (2/3) * n(O2)
n(KClO3) = (2/3) * 0.0948 mol
n(KClO3) = 0.0632 mol

However, the answer choices are given in terms of KClO3, so we need to convert the answer to moles of KClO3:

n(KClO3) = 0.0632 mol * (1 mol KClO3 / 1 mol O2)
n(KClO3) = 0.0632 mol

Therefore, the answer is choice C, 0.0948 mol.

Oxygen Gas, Generated by the Reaction 2KClO3(s) →\rightarrow→ 2KCl(s) + 3O2(g), Is Collected Over Water at 27°C in 3O2(g)

Oxygen Gas, Generated by the Reaction 2KClO₃(s) $\rightarrow$ 2KCl(s) + 3O₂(g), Is Collected Over Water at 27°C in 3O₂(g)