Question
If 43.0 mL of 0.220 M HCl solution is needed to neutralize a solution of Ca(OH)2, how many grams of Ca(OH)2 must b Express the mass in grams to three significant digits.
Answers
GPT-4o mini
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:
\[
\boxed{0.350} \, \text{g}
\]
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:
\[
\boxed{0.350} \, \text{g}
\]