Consider the following task as it relates to the reaction:

A 25.3 g sample of Mg(OH)2 reacts with 22.5 g of HCl and is represented by the equation

Mg(OH)2 + 2 HCl → MgCl2 + 2 H2O

Students must calculate how much MgCl2 is produced given the specific amounts of reactants.

How do you know that this is a limiting reactant problem?
A) The amounts of both reactants are given.
B) The mole ratio of the products is not the same.
C) The mole ratios of the reactants are not the same.
D) There are fewer grams of HCl than Mg(OH)2.

                     D) There are fewer grams of HCl than Mg(OH)₂

Explanation:

                    The balance chemical equation for given reaction is as;

                        Mg(OH)₂ + 2 HCl → MgCl₂ + 2 H₂O

Step 1: Calculate Moles of Mg(OH)₂ and HCl as;

Mg(OH)₂:

              Moles  =  Mass / M.Mass

              Moles  =  25.3 g / 58.319 g/mol

              Moles  =  0.433 moles of Mg(OH)₂

HCl:

              Moles  =  Mass / M.Mass

              Moles  =  22.5 g / 36.460 g/mol

              Moles  =  0.617 moles of HCl

Step 2: Find out Limiting reagent as;

According to equation,

                   1 mole of Mg(OH)₂ reacted with  =  2 moles of HCl

So,

             0.433 moles of Mg(OH)₂ will react with  =  X moles of HCl

Solving for X,

                     X  =  0.433 mol × 2 mol / 1 mol

                    X =  0.866 moles of HCl

This means for given amount of Mg(OH)₂ we require 0.866 moles of Hcl while, we are only provided with 0.617 moles of HCl hence, HCl is the limiting reagent and will control the final yields of products.

Step 3: Find out Moles of MgCl₂ produced;

According to equation,

                   2 moles of HCl produced  =  1 mole of MgCl₂

So,

             0.617 moles of HCl will produce  =  X moles of MgCl₂

Solving for X,

                     X  =  0.617 mol × 1 mol / 2 mol

                    X =  0.3085 moles of MgCl₂

Step 4: Calculate Mass of MgCl₂ as;

                   Mass  =  Moles × M.Mass

                   Mass  =  0.3085 mol × 36.46 g/mol

                   Mass =  11.24 g of MgCl₂