a. CaCO3 + EDTA4- → Ca(EDTA)-2 + CO32-
V(EDTA) = 48.0 mL
Proportion 1:
1000 mL – 0.05 mol
48.0 mL – n mol
n(EDTA) "= \\frac{48.0 \\times 0.05}{1000} = 0.0024\\;mol"
n(CaCO3) = n(EDTA) = 0.0024 mol
M(CaCO3) = 100.08 g/mol
m(CaCO3) "= 0.0024 \\times 100.08 = 0.24 \\;g"
Proportion 2:
1.00 g – 100 %
0.24 – x
x "= \\frac{0.24 \\times 100}{1} = 24 %" %
The percent calcite (CaCO3) content of rock A is 24 %.
b. Mg(OH)2 + EDTA4- → Mg(OH)2EDTA4-
V(EDTA) = 76.5 mL
Proportion 1:
1000 mL – 0.05 mol
76.5 mL – n mol
n(EDTA) "= \\frac{76.5 \\times 0.05}{1000} = 0.003825\\;mol"
n(Mg(OH)2) = n(EDTA) = 0.003825 mol
M(Mg(OH)2) = 58.32 g/mol
m(Mg(OH)2) "= 0.003825 \\times 58.32 = 0.223 \\;g"
Proportion 2:
1.00 g – 100 %
0.223 – x
x "= \\frac{0.223 \\times 100}{1} = 22.3 %" %
The percent brucite (Mg(OH)2) content of rock A is 22.3 %.
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