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Sure, let's work through the solution step by step to determine the mass percentage of calcium hydroxide (Ca(OH)₂) present in the sample.
### Step 1: Calculate the average volume of HCl used in titration
The titration volumes for the three replicates are:
- 35.25 mL
- 35.78 mL
- 35.49 mL
To find the average volume, we sum these volumes and divide by the number of replicates (3):
[tex]\[ \text{Average volume of HCl} = \frac{35.25 + 35.78 + 35.49}{3} \][/tex]
[tex]\[ = \frac{106.52}{3} \][/tex]
[tex]\[ = 35.5067 \, \text{mL} \][/tex]
### Step 2: Calculate the moles of HCl used
The concentration of HCl is given as 0.1541 M, and the average volume of HCl used is 35.5067 mL (which is 0.0355067 L). Thus, the moles of HCl can be found using the formula:
[tex]\[ \text{Moles of HCl} = \text{Concentration} \times \text{Volume} \][/tex]
[tex]\[ = 0.1541 \, \text{M} \times 0.0355067 \, \text{L} \][/tex]
[tex]\[ = 0.0054716 \, \text{mol} \][/tex]
### Step 3: Calculate the moles of Ca(OH)₂
Using the stoichiometry of the reaction:
[tex]\[ \text{Ca(OH)}_2 + 2\text{HCl} \rightarrow \text{CaCl}_2 + 2\text{H}_2\text{O} \][/tex]
we see that 1 mole of Ca(OH)₂ reacts with 2 moles of HCl. Therefore, the moles of Ca(OH)₂ can be calculated as:
[tex]\[ \text{Moles of Ca(OH)₂} = \frac{\text{Moles of HCl}}{2} \][/tex]
[tex]\[ = \frac{0.0054716}{2} \][/tex]
[tex]\[ = 0.0027358 \, \text{mol} \][/tex]
### Step 4: Calculate the mass of Ca(OH)₂ in the aliquot
The molar mass of Ca(OH)₂ is 74.10 g/mol. Therefore, the mass can be determined by multiplying the moles by the molar mass:
[tex]\[ \text{Mass of Ca(OH)₂} = \text{Moles of Ca(OH)₂} \times \text{Molar mass of Ca(OH)₂} \][/tex]
[tex]\[ = 0.0027358 \, \text{mol} \times 74.10 \, \text{g/mol} \][/tex]
[tex]\[ = 0.2027 \, \text{g} \][/tex]
### Step 5: Calculate the mass percentage of Ca(OH)₂ in the sample
The sample mass for the titration was 21.50 g, and we prepared 500 mL of solution from this mass. Since only 5.00 mL of this solution was titrated, we need to account for this by scaling up the mass of Ca(OH)₂ to the entire 500 mL solution. The factor for scaling up is:
[tex]\[ \text{Factor} = \frac{500 \, \text{mL}}{5 \, \text{mL}} = 100 \][/tex]
Therefore, the total mass of Ca(OH)₂ in the original 21.50 g sample is:
[tex]\[ \text{Total mass of Ca(OH)₂} = 0.2027 \, \text{g} \times 100 \][/tex]
[tex]\[ = 20.2721 \, \text{g} \][/tex]
Finally, the mass percentage of Ca(OH)₂ in the sample is:
[tex]\[ \text{Mass percentage of Ca(OH)₂} = \frac{20.2721 \, \text{g}}{21.50 \, \text{g}} \times 100\% \][/tex]
[tex]\[ = 94.29\% \][/tex]
Therefore, the mass percentage of calcium hydroxide in the sample is approximately 94.29%.
### Step 1: Calculate the average volume of HCl used in titration
The titration volumes for the three replicates are:
- 35.25 mL
- 35.78 mL
- 35.49 mL
To find the average volume, we sum these volumes and divide by the number of replicates (3):
[tex]\[ \text{Average volume of HCl} = \frac{35.25 + 35.78 + 35.49}{3} \][/tex]
[tex]\[ = \frac{106.52}{3} \][/tex]
[tex]\[ = 35.5067 \, \text{mL} \][/tex]
### Step 2: Calculate the moles of HCl used
The concentration of HCl is given as 0.1541 M, and the average volume of HCl used is 35.5067 mL (which is 0.0355067 L). Thus, the moles of HCl can be found using the formula:
[tex]\[ \text{Moles of HCl} = \text{Concentration} \times \text{Volume} \][/tex]
[tex]\[ = 0.1541 \, \text{M} \times 0.0355067 \, \text{L} \][/tex]
[tex]\[ = 0.0054716 \, \text{mol} \][/tex]
### Step 3: Calculate the moles of Ca(OH)₂
Using the stoichiometry of the reaction:
[tex]\[ \text{Ca(OH)}_2 + 2\text{HCl} \rightarrow \text{CaCl}_2 + 2\text{H}_2\text{O} \][/tex]
we see that 1 mole of Ca(OH)₂ reacts with 2 moles of HCl. Therefore, the moles of Ca(OH)₂ can be calculated as:
[tex]\[ \text{Moles of Ca(OH)₂} = \frac{\text{Moles of HCl}}{2} \][/tex]
[tex]\[ = \frac{0.0054716}{2} \][/tex]
[tex]\[ = 0.0027358 \, \text{mol} \][/tex]
### Step 4: Calculate the mass of Ca(OH)₂ in the aliquot
The molar mass of Ca(OH)₂ is 74.10 g/mol. Therefore, the mass can be determined by multiplying the moles by the molar mass:
[tex]\[ \text{Mass of Ca(OH)₂} = \text{Moles of Ca(OH)₂} \times \text{Molar mass of Ca(OH)₂} \][/tex]
[tex]\[ = 0.0027358 \, \text{mol} \times 74.10 \, \text{g/mol} \][/tex]
[tex]\[ = 0.2027 \, \text{g} \][/tex]
### Step 5: Calculate the mass percentage of Ca(OH)₂ in the sample
The sample mass for the titration was 21.50 g, and we prepared 500 mL of solution from this mass. Since only 5.00 mL of this solution was titrated, we need to account for this by scaling up the mass of Ca(OH)₂ to the entire 500 mL solution. The factor for scaling up is:
[tex]\[ \text{Factor} = \frac{500 \, \text{mL}}{5 \, \text{mL}} = 100 \][/tex]
Therefore, the total mass of Ca(OH)₂ in the original 21.50 g sample is:
[tex]\[ \text{Total mass of Ca(OH)₂} = 0.2027 \, \text{g} \times 100 \][/tex]
[tex]\[ = 20.2721 \, \text{g} \][/tex]
Finally, the mass percentage of Ca(OH)₂ in the sample is:
[tex]\[ \text{Mass percentage of Ca(OH)₂} = \frac{20.2721 \, \text{g}}{21.50 \, \text{g}} \times 100\% \][/tex]
[tex]\[ = 94.29\% \][/tex]
Therefore, the mass percentage of calcium hydroxide in the sample is approximately 94.29%.
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