Question

Calculate the mass of MgCO₃ precipitated by mixing 10.00 mL of a 0.200 M Na₂CO₃ solution with 5.00 mL of a 0.0550 M Mg(NO₃)₂solution.

Answer 1

Balanced Reaction

Na2CO3 + Mg(NO3)2 → MgCO3 + 2 NaNO3

mmol Na2CO3 = molarity * volume in mL = (10.0 mL) x (0.200 M Na2CO3) = 2.00 mmol

mmol of Mg(NO3)2 = (5.00 mL) x (0.0550 M Mg(NO3)2) = 0.275 mmol

0.275 millimole of Mg(NO3)2 would react completely with 2 millimole of Na2CO3, but since there is Na2CO3 is left hence Na2CO3 is in excess and therefore Mg(NO3)2 is the limiting reactant.

(0.275x 10^-3 mol Mg(NO3)2) x (1 mol MgCO3 / 1 mol Mg(NO3)2) x (84.3139 g MgCO3/mol) =

0.02318 g = 23.18 mg MgCO3

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