Question

Calculate the mass(g.) of MgCO₃ precipitated by mixing 10.00 mL of a 0.200 M Na₂CO₃ solution with 5.00 mL of a 0.0300 M Mg(NO₃)₂solution.

Answer 1

The balanced equation is : Na₂CO₃ (aq) + Mg(NO₃)₂ (aq) MgCO₃ (s) + 2 NaNO₃ (aq)

moles Na₂CO₃ = (molarity Na₂CO₃) * (volume Na₂CO₃ in Liter)

moles Na₂CO₃ = (0.200 M) * (10.00 x 10^-3 L)

moles Na₂CO₃ = 2.00 x 10^-3 mol

moles Mg(NO₃)₂ = (molarity Mg(NO₃)₂) * (volume Mg(NO₃)₂ in Liter)

moles Mg(NO₃)₂ = (0.0300 M) * (5.00 x 10^-3 L)

moles Mg(NO₃)₂ = 1.50 x 10^-4 mol

Since moles of Mg(NO₃)₂ are less than moles of Na₂CO₃, therefore, Mg(NO₃)₂ is the limiting reactant.

moles MgCO₃ formed = moles Mg(NO₃)₂

moles MgCO₃ formed = 1.50 x 10^-4 mol

mass MgCO₃ formed = (moles MgCO₃ formed) * (molar mass MgCO₃)

mass MgCO₃ formed = (1.50 x 10^-4 mol) * (84.3 g/mol)

mass MgCO₃ formed = 0.0126 g