An unknown amount of barium nitrate is dissolved in 120.0 g of water and 8.15 g of copper (II) sulfate is dissolved in 75.0 g of water. When these solutions are mixed 1.076 g of white solid is isolated as a precipitate by vacuum filtration. The filtrate is found to have a mass of 204.44 g. What mass of barium nitrate was dissolved in the water and used for this reaction

The amount of dissolved barium nitrate is 2.366 g

Explanation:

let the amount of dissolved barium nitrate = x

The reaction is given as;

Before reaction:

Mass of barium nitrate solution = (x + 120) g

Mass of copper sulphate solution = (8.15 + 75)g

After reaction:

mass of residue = 1.076 g

mass of filtrate = 204.44 g

Apply the law of conservation of energy;

Total mass before reaction = total mass after reaction

(x + 120) + (8.15 + 75) = 1.076 + 204.44

x + 203.15 = 205.516

x = 205.516 - 203.15

x = 2.366 g

Therefore, the amount of dissolved barium nitrate is 2.366 g