Question

Given the reaction 2HI(aq)+CaCO3(s)=CaI2(aq)+CO2(g)+H2O(l)

What volume of CO2(g) can be produced from 255mL of 3.0M HI and 75.2g of CaCO3 at STP?

Answer 1

2HI(aq) + CaCO₃(s) ----> CaI₂(aq) + CO₂(g) + H₂O(l)

Molar mass of CaCO₃ = 100 g/mole

Thus, moles of CaCO₃in 75.2 g of it = mass/molar mass = 0.752

Moles of HI = molarity*volume of solution in litres = 3*0.255 = 0.765

As per the balanced reaction, HI & CaCO₃ reacts in the molar ratio of 2:1

Thus, CaCO₃ is in excess and HI is the limiting reagent

Hence, Moles of CO₂ produced = (1/2)*moles of HI reacting = 0.3825

Now, at STP, 1 mole of a gas occupies 22.4 litres

Thus, 0.3825 moles of CO₂ at STP requires 22.4*0.3825 = 8.568 litres