Answer to Question #85302 in General Chemistry for Hardin

Question

if 51.23 g of lithium reacts with 402.79 g of iron(III) sulfate, how many grams of iron would you expect to form?

Answer 1

6Li +Fe₂(SO₄)₃ = 3Li₂SO₄ + 2 Fe

n(Li) = m/M = 51.23/6.94 = 7.38 mol

n(Fe₂(SO₄)₃) =m/M =402.79/399.88 = 1.01 mol

Find limiting reactant:

n(Li)/6 = 7.38/6 = 1.23

n(Fe₂(SO₄)₃)/1 = 1.01/1 = 1.01

1.01<1.23 , Fe₂(SO₄)₃ is a limiting reactant

According to equation 1 mol of Fe₂(SO₄)₃ gives 2 mol of Fe

We have 1.01 mol of Fe₂(SO₄)₃ which give x mol of Fe

1/1.01 = 2/x

x = 2.02 mol

n(Fe) = 2.02 mol

m(Fe) = n*M = 2.02* 55.85 = 112.82 g

Answer: 112.82 g