For the following reaction, 15.8 grams of chlorine gas are allowed to react with 8.99 grams of water.

Question

chlorine (g) + water (l) --- hydrochloric acid (aq) + chloric acid (HClO3)(aq)

What is the maximum amount of hydrochloric acid that can be formed?

Mass =  g

What is the FORMULA for the limiting reactant?

What amount of the excess reactant remains after the reaction is complete?

Mass =  g

Answer 1

3Cl₂(g) + 3H₂O(l) ==> 5HCl(aq) + HClO₃(aq) .. balanced equation

Any time you are given the amounts of BOTH reactants, you must first find which reactant is limiting. One way to do this is to divide moles of each reactant by the corresponding coefficient in the balanced equation.

For Cl2: 15.8 g Cl2 x 1 mol Cl2 / 70.2 g = 0.225 mols Cl2 (÷3->0.075)

For H2O: 8.99 g H2O x 1 mol H2O / 18.0 g = 0.499 mols H2O (÷3->0.166)

Since 0.075 is < than 0.166, Cl2 is the limiting reactant, and the moles of Cl2 (0.225 mols) will be used

Use mols of limiting reactant (0.225) to determine amount of HCl that can be formed:

0.225 mols Cl2 x 5 mols HCl / 3 mols Cl2 = 0.375 mols HCl

mass of HCl formed = 0.375 mols HCl x 36.5 g / mol = 13.7 g HCl

To find amount of H2O (excess reactant) left over, find out how much was used, and then deduct that from the amount we started with.

moles H2O used = 0.225 mols Cl2 x 3 mols H2O / 3 mol Cl2 = 0.225 mols H2O used

moles H2O left over = 0.499 mols - 0.225 mols = 0.274 mols H2O left over

mass H2O left over = 0.274 mols H2O x 18.0 g / mol = 4.93 g H2O left over