What is the maximum amount of iron(II) chloride that can be formed?

Question

For the following reaction, 18.7 grams of iron are allowed to react with 28.8 grams of hydrochloric acid.

iron (s) + hydrochloric acid (aq) iron(II) chloride (aq) + hydrogen (g)

What is the maximum amount of iron(II) chloride that can be formed?

What is the FORMULA for the limiting reagent?

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

Answer 1

You really always need to write the correctly balanced equation FIRST, before answering these types of questions. You MUST look at the mole ratio of the reactants and the products.

Fe(s) + 2HCl(aq) ==> FeCl₂(aq) + H₂(g) ... balanced equation (note mole ratios)

First, we must find the limiting reactant:

For Fe(s): 18.7 g Fe x 1 mol Fe/55.8 g x 1 mol FeCl₂/mol Fe x 127 g/mol FeCl₂ = 42.6g FeCl₂ formed

For HCl(aq): 28.8 g HCl x 1 mol HCl/36.5 g x 1 mol FeCl₂/2 mol HCl x 127 g/mol FeCl₂ = 50.1 g FeCl₂ formed

LIMITING REACTANT IS Fe(s)

The maximum amount of FeCl₂ formed is that obtained from the limiting reactant = 42.6 g FeCl₂ formed

Excess reactant (HCl) left: we started with 28.8 g x 1 mol/36.5 g = 0.789 moles. For each mol of Fe, we used 2 moles of HCl. The moles of Fe used is 18.7 g x 1 mol/55.8 g = 0.335 moles Fe. So moles of HCl used is

2 x 0.335 = 0.670 moles HCl used (SEE WHY MOLE RATIOS ARE IMPORTANT?)

Moles HCl left over = 0.789 mol - 0.670 = 0.119 moles HCl left over

mass HCl left over = 0.119 moles HCl x 36.5 g/mol = 4.34 g HCl left over

Answer 2

Caitlyn

Use the same method as for the similar problem you have posted which I answered

Mike