Question

Assuming an efficiency of 36.30%, calculate the actual yield of magnesium nitrate formed from 139.9 g of magnesium and excess copper(II) nitrate. Mg+Cu(N0^3^)2-->Mg(NO^3)^2+Cu

Answer 1

Mg + Cu(N0₃)₂ Mg(NO₃)₂ + Cu

This reaction is equimolar. 1 mole each Mg and Cu(N0₃)₂ react to form 1 mole of Mg(NO₃)₂.

Cu(N0₃)₂ is in excess.

Number of moles of Mg(NO₃)₂ = Mass / Molar Mass = 139.9 / 24.305 = 5.756 moles

Since Cu(N0₃)₂ is in excess, all moles of Mg(NO₃)₂ will react.

Efficiency or Percentage yield = 36.30 %

% yield x theoritical yield / 100 = actual yield

Actual yield = 36.30 x 5.756 / 100 = 2.089 moles of Mg(NO₃)₂

Yield of Mg(NO₃)₂ = number of moles x molar mass = 2.089 x 148.315 = 309.89 g of Mg(NO₃)₂