3K2CrO4 + 2FeBr3 = Fe2(Cr04)3 + 6KBr The 30 g of Iron (III) bromide that reacted could make 20.3 g of Fe2(CrO4)3. What is the limiting reagent in this reaction?

The limiting reagent is FeBr3.

Explanation:

from the reaction, you find a ratio between FeBr3 and Fe2(CrO4)3.

This ratio is 2 moles of FeBr3 reacts and forms 1 mole of Fe2(CrO4)3.

Transform 30 g of FeBr3 in grams:

30 / 296.6 = 0.10 mole

Transform 20.3 g of Fe2(CrO4)3 in grams:

20.3 / 459.7 = 0.0442

You have an excess of Fe2(CrO4)3, the limiting reagent is FeBr3.

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