Asked by skye

Assuming an efficiency of 30.80%, calculate the actual yield of magnesium nitrate formed from 147.4 g of magnesium and excess copper(II) nitrate.
Mg+Cu(NO3)2=Mg(NO3)2+Cu
Answered by skye
is anyone able to answer this?
Answered by oobleck
how many moles of Mg do you have?
The equation says you will get that many moles of Mg(NO3)2
But you will really only get 0.3080 times that number of moles
Now convert that to grams.
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