Question

If 42.7 mL of a 0.208 M HCl solution is needed to neutralize a solution of Ca(OH)2, how many grams of Ca(OH)2 must be in the solution? Ca(OH)2(aq) + 2HCl  CaCl2(aq) + 2H2O(l) g

Answer 1

Answer: The mass of calcium hydroxide that are present in the solution is 0.3256 grams.

Explanation:

To calculate the moles of a solute, we use the equation:

$\text{Molarity of the solution}=\frac{\text{Moles of solute}}{\text{Volume of solution (in L)}}$

We are given:

Volume of hydrochloric acid = 42.7mL = 0.0427 L   (Conversion factor: 1 L = 1000 mL)

Molarity of the solution = 0.208 moles/ L

Putting values in above equation, we get:

$0.208mol/L=\frac{\text{Moles of hydrochloric acid}}{0.0427L}\\\\\text{Moles of hydrochloric acid}=0.0088mol$

For the given chemical reaction:

$Ca(OH)_2(aq.)+2HCl\rightarrow CaCl_2(aq.)+2H_2O(l)$

By Stoichiometry of the reaction:

2 moles of hydrochloric acid reacts with 1 mole of calcium hydroxide.

So, 0.0088 moles of hydrochloric acid will react with = $\frac{1}{2}\times 0.0088=0.0044mol$ of calcium hydroxide.

To calculate the mass of calcium hydroxide, we use the equation:

$\text{Number of moles}=\frac{\text{Given mass}}{\text{Molar mass}}$

Moles of calcium hydroxide = 0.0044 moles

Molar mass of calcium hydroxide = 74 g/mol

Putting values in above equation, we get:

$0.0044mol=\frac{\text{Mass of calcium hydroxide}}{74g/mol}\\\\\text{Mass of calcium hydroxide}=0.3256g$

Hence, the mass of calcium hydroxide that are present in the solution is 0.3256 grams.