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?

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:

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.

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:

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:

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:

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.

1. for the first question, we must find the mass of the anhydrous salt, mgso₄. the molar mass for mgso₄·7h₂o is 246.47 g/mol, while that of mgso₄ is 120.37 g/mol.7.834 g mgso₄·7h₂o * 1 mol mgso₄·7h₂o/246.47 g * 1 mol mgso₄/1 mol mgso₄·7h₂o * 20.37 g mgso₄/mol = 0.647 g mgso₄

2. mass of water = mass of sample - mass of anhydrous salemass of water = 7.834 - 0.647 = 7.19 gpercent water in hydrate = 7.19/7.834 * 100 = 91.74%