What mass of barium sulfate can be produced when 250.0 mL of a 0.400 M  solution of barium chloride is mixed with 250.0 mL of a 0.400 M  solution of iron(III) sulfate? Mass =  g

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justaguide | College Teacher | (Level 2) Distinguished Educator

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The chemical equation for the reaction between barium chloride and iron(III) sulfate is:

`Fe_2(SO4)_3 + 3 BaCl_2 = 2 FeCl_3 + 3 BaSO_4`

When 250.0 mL of a 0.400 M  solution of barium chloride is mixed with 250.0 mL of a 0.400 M  solution of iron(III) sulfate, the limiting reagent is barium chloride.

250 mL of 0.4 M solution of barium chloride has .4*(1/4) = 0.1 mole of barium cloride. Each mole of barium chloride leads to the creation of one mole of barium sulfate. The mass of barium sulfate is 233.43 g/mol. The mass of 0.1 mole is 23.343 g

The mass of barium sulfate produces when 250.0 mL of a 0.400 M  solution of barium chloride is mixed with 250.0 mL of a 0.400 M  solution of iron(III) sulfate is 23.343 g.

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