Question

If 36 g of Cobalt (II) chloride is used in the reaction with 24 g of Mercury (II) phosphate, how much Cobalt (II) phosphate can be produced?

What is the limiting reactant?

Answer 1

This one was a bit tricky, but I gave it a shot anyways.

Cobalt chloride=CoCl2
Mercury phosphate=Hg3(PO4)2
Cobalt phosphate=Co3(PO4)2

Your equation is basically

CoCl2+Hg3(PO4)2>>>>Co3(PO4)2

From there you make two ratios

one for the problem

which will be the moles of Cocl2 and the moles of Hg3(Po4)2

that will be 0.28/0.03=9.3

the equation ratio is just 1:1 in this case

the problem ratio>equation ratio...so the demoninator is limiting and the numerator is excess

excess=Cocl2
limiting=Hg3(PO4)2

now to get the mass for Cobalt (II) phosphate.

take the limiting reactant from the problem, multiply it by the excess in the equation, and the molar mass of Cobalt (II) phosphate

0.28 x 1 x366=roughly 102.48 grams of Cobalt (II) phosphate

Answer 2

That was impressive, thank you, I get it now, I have another one for you if you are up to it. Just watch the feed.

Answer 3

I'll try to get to it.