2.4 mol Fe(OH)3 and 7.9 mol H2SO4 react according to the equation 2Fe(OH)3+3H2SO4−→ Fe2(SO4)3 +6H2O. If the limiting reactant is Fe(OH)3, determine the amount of excess reactant that remains.

Question

2.4 mol Fe(OH)3 and 7.9 mol H2SO4 react according to the equation 2Fe(OH)3+3H2SO4−→ Fe2(SO4)3 +6H2O. If the limiting reactant is Fe(OH)3, determine the amount of excess reactant that remains.

Answer 1

Looking at the balanced equation:

2Fe(OH)₃ + 3H₂SO₄ ==> Fe₂(SO₄)₃ + 6H₂O ... balanced equation

Since you are told that Fe(OH)₃ is the limiting reactant, we will determine how many moles of H₂SO₄ react and then determine how much H2SO4 is left over at the end of the reaction.

moles H2SO4 used: 2.4 mol Fe(OH)₃ x 3 mol H₂SO₄ / 2 mol Fe(OH)₃ = 3.6 moles H₂SO₄ reacted

moles H₂SO₄ remaining = initial - used = 7.9 moles - 3.6 moles = 4.3 moles H₂SO₄ remaining

mass H₂SO₄ remaining = 4.3 moles H2SO4 x 98 g/mol = 421 g H₂SO₄ remaining