Oxygen gas can be prepared in the laboratory by heating potassium chlorate, so that it decomposes according to the equation: 2KClO₃(s) $\rightarrow$ 2KCl(s) + 3O₂(g) What would the pressure of O₂ be in a 2.50 L glass container at 22.0^oC if 6.50 g of KClO₃ is decomposed?

Question

Quizplus · Accepted Answer

First, we need to calculate the moles of O2 produced from the given amount of KClO3:

m(KClO3) = 6.50 g
M(KClO3) = 122.55 g/mol
n(KClO3) = m(KClO3)/M(KClO3) = 0.053 mol

From the balanced equation, we can see that 3 mol of O2 are produced for every 2 mol of KClO3 decomposed, so:

n(O2) = 3/2 * n(KClO3) = 0.0795 mol

Next, we can use the ideal gas law to calculate the pressure of O2 in the container:

V = 2.50 L
n = 0.0795 mol
T = 22.0°C = 295 K
R = 0.08206 L∙atm/(mol∙K)

P = nRT/V = (0.0795 mol)(0.08206 L∙atm/(mol∙K))(295 K)/(2.50 L) = 0.771 atm

Therefore, the correct answer is D) 0.771 atm.

Oxygen Gas Can Be Prepared in the Laboratory by Heating →\rightarrow→

Oxygen Gas Can Be Prepared in the Laboratory by Heating $\rightarrow$