Question

Assuming an efficiency of 31.00%, calculate the actual yield of magnesium nitrate formed from 127.3 g of magnesium and excess copper(II) nitrate.

Mg + Cu(NO3)2 ⟶ Mg(NO3)2 + Cu

Answer 1

Moles of $Mg=\frac{127.3}{24}=5.30mol$

So according to the given efficiency moles of $Mg(NO_3)_2$ formed$=$ $0.31\times5.30=1.643mol$

Hence mass of $Mg(NO_3)_2$ formed$=1.643\times148=243.164g$