Stoichiometry help #9

Question

Given the reactant amounts specified in each chemical equation, determine the amount in moles of excess reactant that remains.

(a) HCl + NaOH —> NaCl + H₂O

2.0 mol HCl and 3.9 mol NaOH

amount of excess reactant

__ mol

(b) Zn + 2 HCl —> ZnCl₂ + H₂

3.3 mol Zn and 4.0 mol HCl

amount of excess reactant

__ mol

(c) 2 Fe(OH)₃ + 3 H₂SO₄ —> Fe₂(SO₄)₃ + 6 H₂O

2.0 mol Fe(OH)₃ and 7.5 mol H₂SO₄

amount of excess reactant

__ mol

Answer 1

This is a problem dealing with LIMITING REACTANTS. I'll do the 3rd one (the most difficult one), and leave it to you to do the others. The procedure / process will be the same for all.

2 Fe(OH)₃ + 3 H₂SO₄ —> Fe₂(SO₄)₃ + 6 H₂O

2.0 mol (Fe(OH)₃ and 7.5 mol H₂SO₄

(step 1) determine which reactant is limiting. easy way to do this is to divide the moles of each reactant by the corresponding coefficient in the balanced equation. Whichever value is less is the limiting reactant.

For Fe(OH)₃: 2.0 mol Fe(OH)₃ ÷ 2->1.0

For H₂SO₄: 7.5 mol H2SO4 ÷ 3->2.5

So, Fe(OH)₃ is THE LIMITING REACTANT

This means that H₂SO₄ is THE REACTANT IN EXCESS

(step 2) determine how much of the excess reactant (H₂SO₄) has been used up. To do this, use the moles of the limiting reactant and the stoichiometry of the equation and dimensional analysis as follows:

2.0 mol Fe(OH)₃ x 3 mols H₂SO₄ / 2 mols Fe(OH)₃ = 3 mols H₂SO₄ used up

(step 3) subtract moles H₂SO₄ used from mols H₂SO₄ initially present to find mols H₂SO₄ left over

7.5 mols H₂SO₄ - 3 mols H₂SO₄ = 4.5 moles H₂SO₄ in excess