Question

If you start with 29.0 grams of Ca(OH)2 and 12.5 grams of HCI how many moles of water can be formed?

Answer 1

The moles of water that will be produced = 0.343 mol

Step-by-step explanation

Given:

Mass of Ca(OH)2 = 29.0 g

Mass of HCl = 12.5 g

We know the reaction : Ca(OH)2 + 2HCI —> CaCI2 + 2H2O

Molar mass of Ca(OH)2 = 74.093 g/mol

Molar mass of HCl = 36.458 g/mol

Molar mass of H2O = 18.015 g/mol

Required: Moles of water that will be formed

Solution:

Use the stoichiometry to find the moles of water using both the reactants.

For Ca(OH)2:

`\begin{gathered} 29.0\text{ g Ca\lparen OH\rparen}_2\text{ x }\frac{1\text{ mole Ca\lparen OH\rparen}_2}{74.093\text{ g Ca\lparen OH\rparen}_2}\text{ x }\frac{2\text{ mol H}_2O}{1\text{ mol Ca\lparen OH\rparen}_2} \\ \\ =\text{ 0.783 mol H}_2O \end{gathered}`

For HCl

`\begin{gathered} 12.5\text{ g HCl x }\frac{1\text{ mol HCl}}{36.458\text{ g}}\text{ x }\frac{2\text{ mol H}_2O}{2\text{ mol HCl}} \\ \\ =\text{ 0.343 mol H}_2O \end{gathered}`

HCl will produce less moles of H2O, thus HCl is the limiting reactant/reagent and the moles of water that will be produced = 0.343 mol