Question

How many grams of iron III sulfate form when 62.0 mL of 0.160 M of Barium...

How many grams of iron III sulfate form when 62.0 mL of 0.160 M of Barium chloride reacts with 58.0 ml of 0.065 M os Sodium Sulfate? Please show the steps necessary to reach the answer as well. This question is not in the textbook solutions section of chegg.

Homework Answers

Answer #1

moles of BaCl2 = 62 x 0.160 / 1000 = 9.92 x 10^-3

moles of Na2SO4 = 58 x 0.065 / 1000 = 3.77 x 10^-3

BaCl2 (aq)     + Na2SO4 (aq) -------------> BaSO4 (s) + 2 NaCl (aq)

   1                    1

9.92 x 10^-3       3.77 x 10^-3

here limiting reagent Na2SO4.

1 mol Na2SO4   --------------> 1 mol BaSO4

3.77 x 10^-3 mol Na2SO4   -----------> ??

moles of BaSO4 formed = 3.77 x 10^-3

mass of BaSO4 formed = 233.39 x 3.77 x 10^-3

                                    = 0.880 g

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