# If 0.370 mol of solid CaCO3 and 940 mL of 0.702 M aqueous H2SO4 are reacted...

 If 0.370 mol of solid CaCO3 and 940 mL of 0.702 M aqueous H2SO4 are reacted stoichiometrically according to the balanced equation, how many moles of solid CaSO4 are produced? H2SO4(aq) + CaCO3(s) → CO2(g) + CaSO4(s) + H2O(l)

molarity of H2SO4 = number of moles of H2SO4 / volume of solution in L

0.702 = number of moles of H2SO4 / 0.940 L

number of moles of H2SO4 = 0.702 * 0.940 = 0.660 mole

from the balanced eqaution we can say that

1 mole of CaCO3 requires 1 mole of H2SO4 so

0.370 mole of CaCO3 will require 0.370 mole of H2SO4

but we have 0.660 mole of H2SO4 so H2SO4 is excess reactant

CaCO3 is limiting reactant

1 mole of CaCO3 produces 1 mole of CaSO4 so

0.370 mole of CaCO3 will produce 0.370 mole of CaSO4

1 mole of CaSO4 = 136.14 g

0.370 mole of CaSO4 = 50.4 g

Therefore, the mass of CaSO4 produced will be 50.4 g

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