- The molar mass of \(\text{Ca(OH)}_2\) is \(74.10 \text{ g/mol}\).
- Calculate the moles of \(\text{Ca(OH)}_2\) as \(n = \frac{5.00}{74.10} \approx 0.06748 \text{ moles}\).
- Determine the required moles of \(\text{HCl}\) using the ratio \(n_{\text{HCl}} = 2 \times n_{\text{Ca(OH)}_2} \approx 0.13496 \text{ moles}\).
- Calculate the volume of \(0.100 \text{ M}\) \(\text{HCl}\) as \(V = \frac{0.13496}{0.100} = \boxed{1.35 \text{ L}}\).
Explanation
- Problem Analysis
The chemical reaction between calcium hydroxide and hydrochloric acid is given by:
\[\text{Ca(OH)}_{2(s)} + 2\text{HCl}_{(aq)} \rightarrow \text{CaCl}_{2(aq)} + 2\text{H}_2\text{O}_{(l)}\]
We are given \(5.00 \text{ grams}\) of calcium hydroxide \(\text{Ca(OH)}_2\) and a hydrochloric acid \(\text{HCl}\) solution with a concentration of \(0.100 \text{ M}\). To determine the volume of \(\text{HCl}\) required for a complete reaction, we must first determine the molar mass of \(\text{Ca(OH)}_2\). Using the atomic masses (Ca \(\approx 40.08\), O \(\approx 16.00\), H \(\approx 1.01\)), the molar mass is \(40.08 + 2 \times (16.00 + 1.01) = 74.10 \text{ g/mol}\).
- Stoichiometric Calculations
Next, we calculate the number of moles of \(\text{Ca(OH)}_2\) present in \(5.00 \text{ grams}\) using the molar mass calculated in the previous step:
\[n_{\text{Ca(OH)}_2} = \frac{\text{mass}}{\text{molar mass}} = \frac{5.00 \text{ g}}{74.10 \text{ g/mol}} \approx 0.06748 \text{ moles}\]
According to the balanced chemical equation, \(1 \text{ mole}\) of \(\text{Ca(OH)}_2\) reacts with \(2 \text{ moles}\) of \(\text{HCl}\). Therefore, the amount of \(\text{HCl}\) needed is:
\[n_{\text{HCl}} = 2 \times n_{\text{Ca(OH)}_2} = 2 \times 0.06748 \approx 0.13496 \text{ moles}\]
- Calculation of Required Volume
Finally, we determine the volume of \(0.100 \text{ M}\) \(\text{HCl}\) solution required. The molarity (\(M\)) is defined as the number of moles per liter (\(M = n / V\)). Thus, the volume \(V\) is:
\[V = \frac{n_{\text{HCl}}}{M} = \frac{0.13496 \text{ moles}}{0.100 \text{ mol/L}} \approx 1.3496 \text{ liters}\]
Rounding this to an appropriate number of significant figures (based on the provided \(5.00 \text{ g}\) and \(0.100 \text{ M}\)), we find that approximately \(1.35 \text{ liters}\) of the hydrochloric acid solution is required.
Examples
Understanding stoichiometric calculations is essential in fields like environmental science and industrial manufacturing. For instance, in water treatment facilities, scientists must precisely calculate the amount of neutralizing agents (like lime, which is Ca(OH)2) to treat acidic runoff or contaminated water before it is safely returned to the environment. This ensures that the water reaches a neutral pH, preventing chemical imbalances in natural water bodies. Calculating the correct molar ratios is the foundational step for both safety and effectiveness in such real-world environmental management tasks.