Given the equation: A 3.00-g sample of KClO₃ is decomposed and the oxygen at 24.0°C and 0.717 atm is collected.What volume of oxygen gas will be collected assuming 100% yield? A) mL B) mL C) mL D) mL E)none of these

Question

Quizplus · Accepted Answer

1 mole of KClO₃ produces 3 moles of oxygen gas according to the balanced chemical equation: 2KClO₃ → 2KCl + 3O2 First, calculate the number of moles of oxygen gas produced: 3 moles O2 / 1 mole KClO₃ x 0.00300 g KClO₃ / 221.3 g/mol KClO₃ = 1.285 x 10^-5 moles O2 Then, use the ideal gas law to calculate the volume of oxygen gas at STP: PV = nRT V = nRT/P V = (1.285 x 10^-5 moles)(0.0821 L·atm/mol·K)(273 K)/(1 atm) V = 0.0279 L = 27.9 mL Therefore, the correct answer is C.

Given the Equation: a 3