What volume of O₂,measured at 24.0°C and 0.882 atm atm,will be produced by the decomposition of 3.00 g KClO₃? Assume 100% yield.(R = 0.08206 L⋅atm/mol⋅K) 2 KClO₃(s)→ 2 KCl(s)+ 3 O₂(g) A) mL B) mL C) mL D) mL E) none of these

Question

Quizplus · Accepted Answer

First, we need to use stoichiometry to find the moles of O2 produced:

3.00 g KClO3 × (1 mol KClO3/122.55 g KClO3) × (3 mol O2/2 mol KClO3) = 0.0361 mol O2

Next, we can use the ideal gas law to find the volume of O2 produced:

PV = nRT

V = (nRT)/P

V = [(0.0361 mol)(0.08206 L⋅atm/mol⋅K)(24.0+273.15 K)]/0.882 atm

V = 0.993 L = 993 mL = 9.93e2 mL

Therefore, the answer is choice A, 11eac9a9_4bd9_9c95_acc3_b328774f9aa5_TB7480_11 mL.

What Volume of O2,measured at 24

What Volume of O₂,measured at 24