What volume of CO2 gas at 645 torr and 800 K could be produced by the reaction of 45 g of CaCO3 according to the equation CaCO3(s) → CaO(s) + CO2(g) (R = 0.08206 L·atm·K-1·mol-1)?
First, determine the moles of CaCO3:
Molar mass of CaCO3 = 40.08 g/mol + 12.01 g/mol + 3(16.00 g/mol) = 100.08 g/mol
Moles of CaCO3 = 45 g / 100.08 g/mol = 0.4494 mol
According to the balanced chemical equation, 1 mol of CaCO3 produces 1 mol of CO2.
Therefore, moles of CO2 produced = 0.4494 mol
Next, calculate the volume of CO2 gas using the ideal gas law equation:
PV = nRT
V = (nRT)/P
V = (0.4494 mol)(0.08206 L·atm·K-1·mol-1)(800 K) / 645 torr
V = 0.0464 L
Therefore, the volume of CO2 gas produced at 645 torr and 800 K from the reaction of 45 g of CaCO3 is 0.0464 L.