a 0.96 millimole sample of Na2CO3 gave 0.9 millimoles of CO2 gas. If a 0.376 g of pure Na2CO3 was reacted with excess acid, what volume of gas will be measured on this apparatus

Answer 1

Na2CO3 + heat ==> Na2O + CO2
Therefore, 1 mol Na2CO3 = 1 mol CO2
Therefore, this apparatus has an efficiency of 0.9/0.96 = 0.9375.

The mols 0.376/molar mass = ?
mols CO2 = the same
Use PV = nRT and solve for volume, then v x 0.9375 since the process is not 100% efficient.