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|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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