Question

57.0mL of a 2.49M solution of HCl react with 25.58g of CaCO3. How any g of CO2 are produced?

2HCl+CaCO3=CO2+H2O+CaCl2

Answer 1

number of moles of HCl = molarity * volume of solution in L

number of moles of HCl = 2.49 * 0.057 = 0.142 mole

Number of moles of CaCO3 = 25.58g / 100.0869 g/mol = 0.2556 mole

from the balanced equation we can say that

2 mole of HCl requires 1 mole of CacO3 so

0.142 mole of HCl will require 0.071 mole of CaCO3

but we have 0.2556 mole of CaCO3 which is in excess so HCl is limiting reactant

2 mole of HCl produces 1 mole of CO2 so

0.142 mole of HCl will produce 0.071 mole of CO2

1 mole of CO2 = 44.01 g

0.071 mole of CO2 = 3.12 g

Therefore, the mass of CO2 produced will be 3.12 g