Question

3. When 1.89 g Al is placed in 50.0 mL of an iron(II) solution, a reaction occurs producing solid metal iron and aluminum ions in solution. After the reaction, the aluminum had a mass of only 1.29 g. How much iron should have been produced? If all the iron reacted, what was the original concentration of the iron(II) solution? If only 1.55 g Fe were produced , what was the percent yield

Answer 1

Let the iron (II) solution be FeSO4

So the reaction will be :

2Al + 3FeSO4 = Al2(SO4)3 + 3Fe

2 moles of Al are produced per three moles of Fe

1.29 g of Al = 1.29 / 26.98 = 0.048 moles

So the number of moles of Fe produced will be : (0.048 x 3) /2 = 0.072 moles

The amount of iron produced = 0.072 x 55.845 = 4.02 g

Concentration of Iron (II) solution = 0.072 / 0.050 = 1.44 M

Percent yield = ( actual yield / theoretical yield) x 100

= ( 1.55 / 4.02) x 100

= 38.56 %