Answer to Question #203812 in Chemistry for Deena Alawdi

Question

For the following reaction, 148 grams of calcium hydroxide are allowed to react with 73.0 grams of hydrochloric acid .

• Ca(OH)₂ (aq) + 2 HCl (aq)  CaCl₂ (aq) + 2 H₂O (l)

What is the FORMULA for the limiting reagent?

What is the maximum amount of calcium chloride that can be formed?  

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

Answer 1

The number of moles of reactants:

n(Ca(OH)₂) = m(Ca(OH)₂) / Mr(Ca(OH)₂) = 148 g / 74 g/mol = 2 mol

n(HCl) = m(HCl) / Mr(HCl) = 73.0 g / 36.5 g/mol = 2 mol

In the reaction, one mole of calcium chloride reacts with two moles of hydrochloric acid. As a result, hydrochloric acid (HCl) is a limiting reagent.

The maximum amount of calcium chloride that can be formed is:

m(CaCl₂) = n(HCl) × Mr(CaCl₂) / 2 = (2 × 111 g/mol) / 2 = 111 g

After the reaction is complete, one mole of Ca(OH)2 remains. From here:

m(Ca(OH)₂) = n(Ca(OH)₂) × Mr(Ca(OH)₂) = 1 mol × 74 g/mol = 74 g