Answer to Question #90459 in Chemistry for Dayanand

Question

What mass of calcium carbonate require to produce 400ml of co2 on heating

Answer 1

CaCO₃ --> CaO + CO₂

According to ideal gas law:

PV=nRT

V_m=n/V=RT/P

Volume of ideal gas at STP is equal to 22.4L for 1 mol.

V_m=22.4 L/mol

n(CO₂)=V(CO₂)/V_m

n(CO₂)=0.4/22.4=0.018mol

n(CO₂):n(CaCO₃)=1:1

n(CaCO₃)=0.018 mol

m(CaCO₃)=n(CaCO₃)*M(CaCO₃)

m(CaCO₃)=0.018*100=1.8g

Answer: m(CaCO₃)=1.8g