J.R. S. answered 12/09/21
Ph.D. University Professor with 10+ years Tutoring Experience
CaCO3 (s) + H2SO4 (aq) ==> CaSO4 + CO2 (g) + H2O (l) ... balanced equation
7.21 g.......1.2L x 0.211M
mols CaCO3 = 7.21 g CaCO3 x 1 mol / 100 g = 0.0721 mols
mols H2SO4 = 1.2 L x 0.211 mol/L = 0.2532 mols
Limiting reactant = CaCO3
mols H2SO4 ued = 0.0721 mols CaCO3 x 1 mol H2SO4 / mol CaCO3 = 0.0721 mols H2SO4 used
mols H2SO4 left over = 0.2532 mols - 0.0721 mol = 0.2261 mols H2SO4 left over
2KOH + H2SO4 ==> K2SO4 + 2H2O ... balanced equation
0.2261 mols H2SO4 x 2 mol KOH / mol H2SO4 = 0.4522 mols KOH needed
0.4522 mols KOH x 1 L / 0.400 mols = 1.13 L KOH needed