(d) If 42.7 mL of 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?

Verified step by step guidance

1. Start by writing the balanced chemical equation for the neutralization reaction between HCl and Ca(OH)2: \[ \text{Ca(OH)}_2 + 2\text{HCl} \rightarrow \text{CaCl}_2 + 2\text{H}_2\text{O} \]. This equation shows that one mole of Ca(OH)2 reacts with two moles of HCl.
2. Calculate the number of moles of HCl used in the reaction. Use the formula \( \text{moles} = \text{concentration} \times \text{volume} \). Convert the volume from mL to L by dividing by 1000: \( 42.7 \text{ mL} = 0.0427 \text{ L} \). Then, calculate the moles: \( 0.208 \text{ M} \times 0.0427 \text{ L} \).
3. Using the stoichiometry from the balanced equation, determine the moles of Ca(OH)2 that reacted. Since 2 moles of HCl react with 1 mole of Ca(OH)2, divide the moles of HCl by 2 to find the moles of Ca(OH)2.
4. Convert the moles of Ca(OH)2 to grams. Use the molar mass of Ca(OH)2, which is approximately \( 74.09 \text{ g/mol} \). Multiply the moles of Ca(OH)2 by its molar mass to find the mass in grams.
5. Review the steps to ensure all calculations are set up correctly and check that units are consistent throughout the process.