Aluminum reacts with 42 g of iron (II) sulfate. How many grams of iron are formed? Aluminum sulfate is the other product?

= 15.44 g

Explanation:

The equation for the reaction;

2Al+3 FeSO4 → Al (2) (SO4) (3) + 3Fe

We can find;

Moles of iron (ii) sulfate = 42 g / 151.908 g/mol

= 0.2765 moles

Mole ratio of iron (ii) sulfate : Iron = 3 : 3 = 1 : 1

Therefore;

Moles of Iron produced = 0.2765 moles

Mass of iron = 0.2765 moles * 55.845 g/mol

= 15.44 g