Question

What mass of magnesium carbonate is precipitated by mixing 10.0 mL of a 0.200 M sodium carbonate solution with 5.00 mL of a 0.0500 M magnesium nitrate solution?

Answer 1

The number of moles of Na2CO3 = vol.in L * mol/L = 10*10^-3 L * 0.2mol/L = 2*10^-3 mol

The number of moles of Mg(NO3)2 = vol.in L * mol/L = 5*10^-3 L * 0.2mol/L = 0.25*10^-3 mol

So the limiting reagent is Mg2+ and

{Q >> Ksp hence ppt will occur.

Q can be calculated with Q = [Mg2+][CO3--] and Ksp = 6.82*10^-6 please note that this is not a part of solution, but it is handy to ensure that ppt is forming or not!!}

Thus the number of moles of MgCO3 formed = 0.25*10^-3 mol

Molar Mass of MgCO3 = 84.31g/mol

Mass of ppt formed = 84.31g/mol * 0.25*10^-3 mol = 0.0211 g