How many mL of O₂ gas at 25°C and 755 mm Hg pressure can be produced from the thermal decomposition of 0.300 grams of KClO₃(s) according to the chemical equation shown below?
2 KClO₃(s) → 2 KCl(s) + 3 O₂(g)

Question

Quizplus · Accepted Answer

First, convert 0.300 g of KClO₃(s) to moles using its molar mass: 0.300 g / 122.54 g/mol = 0.002444 mol According to the balanced chemical equation, 2 moles of KClO₃(s) produces 3 moles of O₂(g) So 0.002444 mol of KClO₃(s) produces: (3 mol O₂(g)/2 mol KClO₃(s)) x 0.002444 mol KClO₃(s) = 0.003666 mol O₂(g) Next, use the ideal gas law to find the volume of O₂(g) produced: PV = nRT where P = 755 mm Hg (convert to atm by dividing by 760), T = 25 + 273.15 = 298.15 K, n = 0.003666 mol, R = 0.08206 L atm/mol K V = (nRT)/P V = (0.003666 mol x 0.08206 L atm/mol K x 298.15 K)/(755 mm Hg/760 mm Hg/atm) V = 0.09039 L = 90.4 mL Therefore, the answer is C: 90.4 mL.

How Many ML of O2 Gas at 25°C and 755

How Many ML of O₂ Gas at 25°C and 755