Question

Solutions of sodium carbonate and silver nitrate react to form solid silver carbonate and a solution of sodium nitrate. A solution containing 3.50 g of sodium carbonate is mixed with one containing 5.00 g of silver nitrate. How many grams of sodium carbonate, silver nitrate, silver carbonate, and sodium nitrate are present after the reaction is complete?

Answer 1

Answer:

1) The no. of grams of Na₂CO₃ = 1.96 g.

2) The no. of grams of AgNO₃ = 0.0 g, because it is the limiting reactant that is consumed completely.

3) The no. of grams of Ag₂CO₃ = 3.998 g ≅ 4.0 g.

4) The no. of grams of NaNO₃ = 2.498 g ≅ 2.5 g.

Explanation:

• For the balanced equation:

Na₂CO₃(aq) + 2AgNO₃(aq) → Ag₂CO₃(s) + 2NaNO₃,

It is clear that 1 mol of Na₂CO₃ and 2 mol of AgNO₃ to produce 1 mol of Ag₂CO₃ and 2 mol of NaNO₃.

• Firstly, we need to calculate the no. of moles of 3.50 g of Na₂CO₃ and 5.00 g of AgNO₃:

no. of moles of Na₂CO₃ = mass/molar mass = (3.5 g)/(105.9888 g/mol) = 0.033 mol.

no. of moles of AgNO₃ = mass/molar mass = (5.0 g)/(169.87 g/mol) = 0.0294 mol.

1) the no. of grams of sodium carbonate, and silver nitrate:

• We need to determine the limiting reactant:

• From stichiometry of the balanced equation:

1 mol of Na₂CO₃ reacts completely with 2 mol of AgNO₃ with (1: 2 molar ratio).

∴ 0.0145 mol of Na₂CO₃ "the excess reactant, and the remaining is in excess (0.033 mol - 0.0145 mol = 0.0185 mol)" reacts completely with (0.0294 mol) of AgNO₃ "limiting reactant".

∴ The no. of grams of Na₂CO₃ = (no. of moles remaining)(molar mass) = (0.0185 mol)((105.9888 g/mol) = 1.96 g.

∴ The no. of grams of AgNO₃ = 0.0 g, because it is the limiting reactant that is consumed completely.

2) the no. of grams of silver carbonate:

• Firstly, we need to find the no. of moles of silver carbonate:

Using cross multiplication:

1 mol of Na₂CO₃ produces → 1 mol of Ag₂CO₃, from stichiometry.

∴ 0.0145 mol of Na₂CO₃ produces → 0.0145 mol of Ag₂CO₃.

∴ The no. of grams of Ag₂CO₃ = (no. of moles of Ag₂CO₃)(molar mass of Ag₂CO₃) = (0.0145 mol)(275.7453 g/mol) = 3.998 g ≅ 4.0 g.

3) the no. of grams of sodium nitrate:

• Firstly, we need to find the no. of moles of sodium nitrate:

Using cross multiplication:

2 mol of AgNO₃ produces → 2 mol of NaNO₃, from stichiometry.

∴ 0.0294 mol of AgNO₃ produces → 0.0294 mol of NaNO₃.

∴ The no. of grams of NaNO₃ = (no. of moles of NaNO₃)(molar mass of NaNO₃) = (0.0294 mol)(84.9947 g/mol) = 2.498 g ≅ 2.5 g.