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) $1.01 \times 10 ^ { 3 }$ mL B) $6.76 \times 10 ^ { 2 }$ mL C) $8.2 \times 10 ^ { 1 }$ mL D) $4.51 \times 10 ^ { 2 }$ mL E) none of these

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Quizplus · Accepted Answer

The balanced chemical equation is:2 KClO3(s) → 2 KCl(s) + 3 O2(g)The molar mass of KClO3 is 122.55 g/mol.The number of moles of KClO3 is:3.00 g / 122.55 g/mol = 0.0245 molAccording to the balanced chemical equation, 2 moles of KClO3 produce 3 moles of O2.Therefore, 0.0245 mol of KClO3 will produce:0.0245 mol * (3 mol O2 / 2 mol KClO3) = 0.0368 mol O2The volume of O2 at 24.0°C and 0.882 atm is:V = nRT/PV = (0.0368 mol) * (0.08206 L?atm/mol?K) * (24.0 + 273.15) K / (0.882 atm)V = 1.01 L = 1010 mLTherefore, the correct answer is A) 1.01 × 10^3 mL.

What Volume of O2, Measured at 24 1.01×1031.01 \times 10 ^ { 3 }1.01×103

What Volume of O₂, Measured at 24 $1.01 \times 10 ^ { 3 }$