Question

Assuming an efficiency of 49.40%, calculate the actual yield of magnesium nitrate formed from 130.8 g of magnesium and excess copper(II) nitrate.

Mg+Cu(NO3)2⟶Mg(NO3)2+Cu

Answer 1

Moles of $Mg$ reacted$=\frac{130.8g}{24\frac{g}{mol}}=5.45mol$

According to the given efficiency, moles of $Mg(NO_3)_2$ formed$=0.4940\times5.45=2.6923mol$

Mass of $Mg(NO_3)_2$ formed$=2.6923mol\times148\frac{g}{mol}=398.4604g$