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.

Question

Mg+Cu(NO3)2=Mg(NO3)2+Cu
is anyone able to solve this? ive gotten the answer wrong twice

Answer 1

Mg + Cu(NO3)2 ==>Mg(NO3)2 + Cu
mols Mg = 147.4/atomic mass Mg = ?
mols Cu(NO3)2 formed = mols Mg from above since it is 1 mol Mg = 1 mol Cu(NO3)2.
g Cu(NO3)2 = mol Cu(NO3)2 x molar mass Cu(NO3)2 = ?
That is the theoretical yield if it were 100%.
Take that number and multiply by 0.3080 = ?
Piece o' cake.