If 125 g of CaCO3 is mixed and reacted with 125 g of HCl, which reactant is limiting and how many grams of CO2 can be made?

CaCO3 is the limiting reactant

55 g of CO2 is made

Explanation:

First we must put down the reaction equation;

CaCO3(s) + 2HCl(aq) > CaCl2(s) + H2O(l) + CO2(g)

Number of mole of CaCO3 = 125g/100gmol-1 = 1.25 moles

From the reaction equation;

1 mole of CaCO3 yields 1 mole of CO2

Hence 1.25 moles of CaCO3 yields 1.25 moles of CO2

For HCl;

number of moles of HCl = 125g/36.5 g mol-1 = 3.42 moles

From the reaction equation;

2 moles of HCl yields 1 mole of CO2

3.42 moles of HCl yields 3.42 * 1/2 = 1.71 moles of CO2

Hence CaCO3 is the limiting reactant.

Mass of CO2 produced = 1.25g * 44 gmol-1 = 55 g of CO2

56 moles of   lieter in chemistry