Question

Assuming an efficiency of 39.20%, calculate the actual yield of magnesium nitrate formed from 133.4 g of magnesium and excess copper (ll)nitrate.

Mg+Cu(NO3)2--------->Mg(NO3)+Cu

Answer 1

Mg + Cu(NO3)2 Mg(NO3) + Cu

First lets balance the equation:

Mg + Cu(NO₃)₂ Mg(NO₃)₂ + Cu

Thus 1 mole of Mg will react with 1 mole of Cu(NO3)2 to give 1 mole of Mg(NO3)2 and limitting reagent is Mg.

133.4 g of Mg = (133.4 g) / 24.305 g/mol = 5.4886 moles.

So, 5.4886 moles of Mg should yield 5.4886 moles of Mg(NO3)2.

5.4886 moles of Mg(NO3)2 = (148.3148 g/mol x 5.4886 mole) = 814.0406 g of Mg(NO3)2

Thus 814.0406 g = 100% theoretical yield

Efficiency is 39.2%. This means actual yield would be = 814.0406 g x 0.392 = 319.1039 g

Actual yiled is = 319.1039 g