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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

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onyebuchinnaji

Answer:

The amount of dissolved barium nitrate is 2.366 g

Explanation:

let the amount of dissolved barium nitrate = x

The reaction is given as;

[tex]Ba(NO_3)_2 _{(aq)} + CuSO_4 _{(aq)} --> BaSO_4 _{(s)} + Cu(NO_3)_2_{(aq)}[/tex]

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

pstnonsonjoku

2.366 g of barium nitrate was dissolved in the water and used for this reaction

The equation of the reaction is;

Ba(NO3)2(aq) + CuSO4(aq) ------> BaSO4(s) + Cu(NO3)2(aq)

From the question, we have the following information;

  • The amount of barium nitrate dissolved in 120.0 grams of water is not known.
  • 8.15 grams of copper (II) sulfate was dissolved in 75.0 grams of water
  • The mass of the precipitate formed thereafter is 1.076 grams
  • The mass of the filtrate is 204.44 grams

From the law of conservation of mass; the total mass before reaction = total mass after reaction.

Let x be the mass of barium nitrate dissolved in water.

(204.44 g + 1.076 g ) = (120.0 g + 75.0 g + 8.15 g + x)

205.516 = 203.15  + x

x = 205.516 - 203.15

x = 2.366 g of barium nitrate

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