Question

385mg of iron reacts with excess bromine, producing 1921mg of a mixture of FeBr2 and FeBr3. How would you calculate the masses of FeBr2 and FeBr3 in the product mixture?

Answer 1

Answer:

Here's what I got.

Explanation:

You know that 385 mg of iron will react with excess bromine to produce mixture of iron(II) bromide,FeBr2,and iron(III) bromide,FeBr3.

The total mass of the mixture is equal to

1921 mg.

Your strategy here will be to write the balanced chemical equations for these two reactions, then use the mole ratios that exist between iron and the two products to determine how many moles of each were produced by the reaction.

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