What volume of CO₂ gas at 645 torr and 800. K could be produced by the decomposition of 45.0 g of CaCO₃? CaCO₃(s) → CaO(s) + CO₂(g)
(R = 0.08206 L • atm/K • mol)

Question

What volume of CO₂ gas at 645 torr and 800. K could be produced by the decomposition of 45.0 g of CaCO₃? CaCO₃(s) → CaO(s) + CO₂(g) (R = 0.08206 L • atm/K • mol)

Quizplus · Accepted Answer

First, calculate the moles of CaCO3 using its molar mass (100.09 g/mol for CaCO3). Moles of CaCO3 = 45.0 g / 100.09 g/mol = 0.449 mol. According to the balanced equation, 1 mole of CaCO3 produces 1 mole of CO2, so 0.449 mol of CaCO3 will produce 0.449 mol of CO2. Use the ideal gas law PV = nRT, converting pressure to atm (645 torr / 760 torr/atm = 0.8487 atm) and solving for V: V = (nRT)/P = (0.449 mol * 0.08206 L·atm/K·mol * 800 K) / 0.8487 atm = 34.8 L.

What Volume of CO2 Gas at 645 Torr and 800

What Volume of CO₂ Gas at 645 Torr and 800