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...

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?

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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

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