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When aqueous solutions of Na2SO4 and Pb(NO3)2 are rnmixed, PbSO4 precipitates. Calculate the mass of PbSO4 (in grams) formed when a 1.25 l solution of rn0.0500 m Pb(NO3)2 and a 2.00 l solution of 0.0250 m Na2SO4 are mixed.
Accepted Answer
1. Calculate the moles of Pb(NO3)2 and Na2SO4 present in the solutions
Moles of Pb(NO3)2:
Moles = Concentration x Volume
Moles of Pb(NO3)2 = 0.0500 mol/L x 1.25 L
Moles of Pb(NO3)2 = 0.0625 mol
Moles of Na2SO4:
Moles = Concentration x Volume
Moles of Na2SO4 = 0.0250 mol/L x 2.00 L
Moles of Na2SO4 = 0.0500 mol
2. Determine the limiting reactant
From the balanced chemical equation:
Pb(NO3)2(aq) + Na2SO4(aq) → PbSO4(s) + 2NaNO3(aq)
The mole ratio of Pb(NO3)2 to Na2SO4 is 1:1. Since we have 0.0625 mol of Pb(NO3)2 and 0.0500 mol of Na2SO4, Pb(NO3)2 is in excess and Na2SO4 is the limiting reactant.
3. Calculate the moles of PbSO4 formed
The stoichiometry of the balanced equation shows that 1 mole of Na2SO4 reacts with 1 mole of Pb(NO3)2 to form 1 mole of PbSO4.
Therefore, moles of PbSO4 formed = moles of Na2SO4
Moles of PbSO4 formed = 0.0500 mol
4. Convert moles of PbSO4 to grams
Mass = Moles x Molar mass
Molar mass of PbSO4 = 303.25 g/mol
Mass of PbSO4 formed = 0.0500 mol x 303.25 g/mol
Mass of PbSO4 formed = 15.16 g
Therefore, when the solutions are mixed, 15.16 g of PbSO4 will precipitate.
Moles of Pb(NO3)2:
Moles = Concentration x Volume
Moles of Pb(NO3)2 = 0.0500 mol/L x 1.25 L
Moles of Pb(NO3)2 = 0.0625 mol
Moles of Na2SO4:
Moles = Concentration x Volume
Moles of Na2SO4 = 0.0250 mol/L x 2.00 L
Moles of Na2SO4 = 0.0500 mol
2. Determine the limiting reactant
From the balanced chemical equation:
Pb(NO3)2(aq) + Na2SO4(aq) → PbSO4(s) + 2NaNO3(aq)
The mole ratio of Pb(NO3)2 to Na2SO4 is 1:1. Since we have 0.0625 mol of Pb(NO3)2 and 0.0500 mol of Na2SO4, Pb(NO3)2 is in excess and Na2SO4 is the limiting reactant.
3. Calculate the moles of PbSO4 formed
The stoichiometry of the balanced equation shows that 1 mole of Na2SO4 reacts with 1 mole of Pb(NO3)2 to form 1 mole of PbSO4.
Therefore, moles of PbSO4 formed = moles of Na2SO4
Moles of PbSO4 formed = 0.0500 mol
4. Convert moles of PbSO4 to grams
Mass = Moles x Molar mass
Molar mass of PbSO4 = 303.25 g/mol
Mass of PbSO4 formed = 0.0500 mol x 303.25 g/mol
Mass of PbSO4 formed = 15.16 g
Therefore, when the solutions are mixed, 15.16 g of PbSO4 will precipitate.
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