a) 3CaCO3 + 2H3PO4 → Ca3(PO4)2 + 3CO2 + 3H2O
b) M(CaCO3) = 100.08 g/mol
n(CaCO3) "= \\frac{12.5}{100.08} = 0.125\\;mol"
M(H3PO4) = 97.99 g/mol
n(H3PO4) "= \\frac{17.3}{97.99} = 0.176 \\;mol"
According to the reaction equation for two moles of H3PO4 we need 3 moles of CaCO3. So, CaCO3 is a limiting reactant.
n(Ca3(PO4)2) "= \\frac{1}{3} n(CaCO_3) = \\frac{1}{3} \\times 0.125 = 0.0416 \\;mol"
M(Ca3(PO4)2) = 310.18 g/mol
m(Ca3(PO4)2) "= 0.0416 \\times 310.18 = 12.924 \\;g"
c) According to the reaction equation for two moles of H3PO4 we need 3 moles of CaCO3. So, CaCO3 is a limiting reactant.
n(Ca3(PO4)2) "= \\frac{1}{3} n(CaCO_3) = \\frac{1}{3} \\times 3.5 = 1.166 \\;mol"
m(Ca3(PO4)2) "= 1.166 \\times 310.18 = 361.876 \\;g"
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