Assuming an efficiency of 39.80%, calculate the actual yield of magnesium nitrate formed from 127.6 g of magnesium and excess copper(II) nitrate.
chemical equation:
Mg(s)+Cu(NO3)2(aq)→Mg(NO3)2(aq)+Cu(s)
One mole of magnesium will react with one mole of copper(II) nitrate to form one mole of magnesium nitrate. 1:1 mole ratios.
molar mass of magnesium=24.305 g
=127.6gx1 mole Mg/ 24.305g
=5.249 moles Mg
Since you have a 1:1 mole ratio
=5.249moles Mgx1 mole Mg(NO3)2/1mole Mg
=5.249 moles Mg
% yield=actual yield/ theoretical yield×100
The actual yield will be equal to
actual yield=% yield×theoretical yield/100
actual yield=39.80x5.249 moles/ 100
=2.089 moles
= 2.089molesx148.315 g/1mole (magnesium nitrate molar mass= 148.315g)
=309.84 g Mg(NO3)2
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