What is the actual yield of magnesium nitrate?

Question

Assuming an efficiency of 37.30%, calculate the actual yield of magnesium nitrate formed from 132.9 g of magnesium and excess copper(II) nitrate.

Mg+Cu(NO3)2 ——-> Mg(NO3)2+Cu

Answer 1

Mg+Cu(NO₃)₂ ——-> Mg(NO₃)₂ + Cu ... balanced equation

Find the moles of magnesium (Mg) used:

132.9 g of Mg x 1 mole Mg/24.305 g = 5.468 moles Mg used

Calculate theoretical yield of magnesium nitrate, Mg(NO3)2:

From the balanced equation, the mole ratio of Mg(NO3)2 : Mg is 1:1

5.468 mol Mg x 1 mol Mg(NO3)2 / mole Mg = 5.468 moles

mass of Mg(NO3)2 = 5.468 moles Mg(NO3)2 x 1 mol/148.3 g = 810.9 g = theoretical yield of Mg(NO3)2

Assuming an efficiency of 37.30%, the actual yield would be 37.30% x 810.9 g = 302.5 g