Verified Answer :
- Calculate the moles of \(HCl\) present: \(n_{HCl} = 0.500\text{ mol/L} \times 0.075\text{ L} = 0.0375\text{ mol}\).
- Determine the moles of \(CaCl_2\) from stoichiometry: \(n_{CaCl_2} = \frac{n_{HCl}}{2} = 0.01875\text{ mol}\).
- Find the molar mass of \(CaCl_2\): \(110.98\text{ g/mol}\).
- Calculate the final mass: \(m_{CaCl_2} = 0.01875\text{ mol} \times 110.98\text{ g/mol} = \boxed{2.08\text{ g}}\).
Explanation
- Analyze the Chemical Equation and Initial Conditions
The chemical reaction between calcium hydroxide, \(Ca(OH)_2\), and hydrochloric acid, \(HCl\), is given by the balanced equation:
\[Ca(OH)_2 + 2HCl \rightarrow CaCl_2 + 2H_2O\]
From this balanced equation, we see that one mole of calcium chloride (\(CaCl_2\)) is produced for every two moles of hydrochloric acid (\(HCl\)) that react. We are given \(75.0\text{ mL}\) of \(0.500\text{ M}\) \(HCl\). First, we convert the volume of the acid from milliliters to liters:
\[V_{HCl} = 75.0\text{ mL} \times \frac{1\text{ L}}{1000\text{ mL}} = 0.075\text{ L}\]
- Calculate the Moles of Reactant and Product
Next, we calculate the number of moles of \(HCl\) using its molarity and volume:
\[n_{HCl} = M \times V = 0.500\text{ mol/L} \times 0.075\text{ L} = 0.0375\text{ moles}\]
Based on the stoichiometry of the reaction, the number of moles of \(CaCl_2\) produced is half the number of moles of \(HCl\) consumed:
\[n_{CaCl_2} = \frac{n_{HCl}}{2} = \frac{0.0375\text{ mol}}{2} = 0.01875\text{ moles}\]
- Determine the Mass of the Product
To find the mass of the produced \(CaCl_2\), we first need its molar mass. Using the atomic masses of calcium (approximately \(40.08\text{ g/mol}\)) and chlorine (approximately \(35.45\text{ g/mol}\)):
\[\text{Molar Mass of } CaCl_2 = 40.08 + 2 \times 35.45 = 110.98\text{ g/mol}\]
Finally, the mass of \(CaCl_2\) formed is:
\[m_{CaCl_2} = n_{CaCl_2} \times \text{Molar Mass} = 0.01875\text{ mol} \times 110.98\text{ g/mol} = 2.080875\text{ grams}\]
Rounding appropriately to three significant figures, we get \(\boxed{2.08\text{ g}}\).
Examples
In industrial chemistry, determining the mass of a product formed is crucial for calculating the yield of a reaction. For example, if a manufacturer wants to produce a specific amount of calcium chloride for road de-icing or dust control, they use stoichiometry like this to figure out exactly how much hydrochloric acid is needed for a batch, ensuring efficiency and minimizing waste. This is a practical application of conservation of mass in a controlled laboratory or factory setting.