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Stoichiometry problem dealing with limiting reagents

If 275.0 mL of 0.5000 M ZnCl2 reacts with 1.650 x 10 to 23rd power of AlPO4 compounds in the following reaction, what is the limiting reagent, how many moles of excess reagent will there be, and how many grams of each product will form?
3 ZnCl2 + 2 AlPO4 yields Zn3(PO4)2 + 2 AlCl3

I am substitute teaching, and I would like a step-by-step solution to be able to post on the board to the students. Thank you.

Solution Preview

BASICS: MOLE CONCEPT.
3 ZnCl2 + 2 AlPO4 yields Zn3(PO4)2 + 2 AlCl3
=> 3 moles of ZnCl2 + 2 moles of AlPO4
= 1 mole of Zn3(PO4)2 + 2 moles of AlCl3

275 ml of 0.5M ZnCl2 == 275*0.5 ml of M ZnCl2
== 275*0.5/1000 mole of ZnCl2 = 0.1375 mole of ZnCl2

1.65*10^(23) molecules of AlPO4
== ...

Solution Summary

The solution is given step-by-step.

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