To determine how many grams of \( \text{Ca(OH)}_2 \) are needed to be neutralized by the \( \text{HCl} \) solution, we need to utilize stoichiometry.
First, let's write the balanced neutralization reaction:
\[ \text{Ca(OH)}_2 + 2\text{HCl} \rightarrow \text{CaCl}_2 + 2\text{H}_2\text{O} \]
From this reaction, we see that one mole of \( \text{Ca(OH)}_2 \) reacts with two moles of \( \text{HCl} \).
Step 1: Calculate the moles of HCl used.
We are given the volume and concentration of the HCl solution:
\[ \text{Volume of HCl} = 43.0 , \text{mL} = 0.0430 , \text{L} \] \[ \text{Concentration of HCl} = 0.220 , \text{M} \]
The number of moles of \( \text{HCl} \) can be calculated as follows:
\[ \text{Moles of HCl} = \text{Concentration} \times \text{Volume} = 0.220 , \text{mol/L} \times 0.0430 , \text{L} = 0.00946 , \text{mol} \]
Step 2: Calculate the moles of Ca(OH)2 neutralized.
From the balanced equation, we know that 1 mole of \( \text{Ca(OH)}_2 \) reacts with 2 moles of \( \text{HCl} \). Thus, we can find the moles of \( \text{Ca(OH)}_2 \):
\[ \text{Moles of Ca(OH)}_2 = \frac{\text{Moles of HCl}}{2} = \frac{0.00946 , \text{mol}}{2} = 0.00473 , \text{mol} \]
Step 3: Calculate the mass of Ca(OH)2.
The molar mass of \( \text{Ca(OH)}_2 \) is calculated as follows:
- Molar mass of Ca = 40.08 g/mol
- Molar mass of O = 16.00 g/mol (2 O in \( \text{Ca(OH)}_2 \))
- Molar mass of H = 1.01 g/mol (2 H in \( \text{Ca(OH)}_2 \))
So, the molar mass of \( \text{Ca(OH)}_2 \) is:
\[ \text{Molar mass of Ca(OH)}_2 = 40.08 , \text{g/mol} + (2 \times 16.00 , \text{g/mol}) + (2 \times 1.01 , \text{g/mol}) = 40.08 , \text{g/mol} + 32.00 , \text{g/mol} + 2.02 , \text{g/mol} = 74.10 , \text{g/mol} \]
Now we can calculate the mass of \( \text{Ca(OH)}_2 \):
\[ \text{Mass of Ca(OH)}_2 = \text{Moles} \times \text{Molar mass} = 0.00473 , \text{mol} \times 74.10 , \text{g/mol} = 0.35004 , \text{g} \]
Step 4: Round to three significant digits.
The mass of \( \text{Ca(OH)}_2 \) rounded to three significant digits is:
\[ \text{Mass of Ca(OH)}_2 \approx 0.350 , \text{g} \]
Thus, the final answer is:
\[ \boxed{0.350} , \text{g} \]