What is the maximum yield in grams of Cl2 that can be produced by reacting 70.0g of KMnO4 with 75.0 g of HCl

Answer 1

2KMnO₄ + 16HCl → 2KCl + 2MnCl₂ + 5Cl₂ + 8H₂O ... balanced equation

If you don't know how to obtain this balanced equation, you can submit another question asking about that.

Since we are given the mass of both reactants, we will have find out which, if either, is limiting. An easy way to do this is to divide the moles of each reactant by its coefficient in the balanced equation, and whichever value is less is the limiting reactant.

For KMnO₄: 70.0 g KMnO₄ x 1 mol KMnO₄ / 158.0 g = 0.4430 mols KMnO₄ (÷2 -> 0.2215)

For HCl: 75.0 g HCl x 1 mol HCl / 36.46 g = 2.057 mols HCl (÷16 -> 0.1286)

Since 0.1286 is less than 0.2215, HCl IS LIMITING

Now we use the 2.057 mols HCl to find the amount of Cl2 that can be formed:

2.057 mols HCl x 5 mols Cl₂ / 16 mols HCl x 70.91 g Cl₂ / mol Cl₂ = 45.58 g Cl₂ = 45.6 g Cl₂ (3 sig.figs.)