Question

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.

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Answer 1

Iron (III) means Fe³⁺

we have reactants

BaCl₂ + Na2SO4  

and product

Fe₂(SO₄)₃

The question seems incorrect ,how can Fe(III)sulfate be formed when Fe is not present in any of the reactants

please check the question once and comment down for the solution

In case Iron(III) chloride + Sodium sulfate is there ,the reaction can be written as:

2FeCl₃ + 3Na₂SO₄ -----------> Fe₂(SO4)₃ + 6NaCl

this can be solved as both side the no. of atoms are balanced