Question

Assuming a yield of 29.70%, calculate the actual yield of magnesium nitrate formed from 133.1 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

First let us calculate theoretical yield of Mg, That means

How much Mg(NO3)2 is formed from 133.1 grams Mg?

grams Mg(NO3)2 =

133.1 grams (Mg) x (1 mol Mg/24.3 g) (Mg) x (1 mol Mg(NO3)2 /1 mol (Mg) x (148.3 g Mg(NO3)2 /1 mol Mg(NO3)2)

All the units except grams Mg(NO3)2 cancel out, leaving

grams Mg(NO3)2 = (133.1 x 1/24.3 x 1 x 148.3) grams H₂O

grams Mg(NO3)2= 812.3

We know the formula
The %yield = (Actual yield/Theoretical yield) *100

let us consider the actual yield of magnesium = X g, then

29.7 = (X / 812.3) *100

X = (29.7 x 812.3) / 100 = 241.3 g