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Assuming an efficiency of 46.80%, calculate the actual yield of magnesium nitrate formed from 147.9g of magnesium and excess copper(II) nitrate.

Mg + Cu(NO3)2 --> Mg(NO3)2 + Cu

I can't figure out how to even set this up much less solve it...

3 answers

for each mole of Mg, you get a mole of magnesium nitrate.
So moles Mg=147.9/atomicmassMg

and you should get the same number of moles of Magnesium nitrate. Convert that to grams by multiplying by molmass of Mg(NO3)2

now for actual yield, multiply by .468
But where did the 147.9 come from?
Read the problem. It says you started with 147.9 g Mg.