Q3.104P - Chemistry: Molecular Nature Of Matter And Change

Question

How many milliliters of 0.383 M HCl are needed to react with 16.2 g of CaCO₃?

$\mathbf{2}\mathit{H}\mathit{C}\mathit{l}\mathbf{\left(}\mathit{a}\mathit{q}\mathbf{\right)}\mathbf{+}\mathit{C}\mathit{a}\mathit{C}{\mathit{O}}_{\mathbf{3}}\mathbf{\left(}\mathit{s}\mathbf{\right)}\mathbf{\to }\mathit{C}\mathit{a}\mathit{C}{\mathit{l}}_{\mathbf{2}}\mathbf{\left(}\mathit{a}\mathit{q}\mathbf{\right)}\mathbf{+}\mathit{C}{\mathit{O}}_{\mathbf{2}}\mathbf{\left(}\mathit{g}\mathbf{\right)}\mathbf{+}{\mathit{H}}_{\mathbf{2}}\mathit{O}\mathbf{\left(}\mathit{l}\mathbf{\right)}$

Step-by-Step Solution

Answer

The amount of solution needed to react with 16.2 g of CaCO₃ is an 846 mL HCl solution.

Step 1: Calculating the number of moles of HCl

Calculate the number of moles of HCl needed to react with 16.2 g of Calcium chloride.

$\begin{array}{rcl}Moles of CaC{O}_3& =& 16.2 g CaC{O}_3\times \frac{1 mol CaC{O}_3}{100.09 g CaC{O}_3}\times \frac{2 mol HCl}{1 mol CaC{O}_3}\\ & =& 0.324 mol HCl.\end{array}$

Step 2: Calculating the volume of solution

Calculate the volume of the solution needed as shown below:

$\begin{array}{rcl}Volume of solution& =& \frac{0.324 mol HCl}{0.383 \frac{mol HCl}{L so\ln }}\\ & =& 0.846 L solution or 846 mL solution.\end{array}$