Sure, let's solve this step by step.
### 7.5.2
To calculate the number of moles of pure potassium hydroxide (KOH) that react with sulfuric acid (H₂SO₄), follow these steps:
1. Determine the purity of KOH:
The potassium hydroxide pellets are 80% pure. This means that out of 100 grams of pellets, 80 grams are pure KOH.
2. Calculate the mass of pure KOH:
Given that we are using 100 grams of potassium hydroxide pellets:
[tex]\[
\text{Mass of pure KOH} = \text{Total mass} \times \text{Purity} = 100 \text{ g} \times 0.80 = 80 \text{ g}
\][/tex]
3. Find the molar mass of KOH:
The molar mass of KOH is given as 56.11 g/mol.
4. Calculate the number of moles of pure KOH:
[tex]\[
\text{Moles of pure KOH} = \frac{\text{Mass of pure KOH}}{\text{Molar mass of KOH}} = \frac{80 \text{ g}}{56.11 \text{ g/mol}} \approx 1.426 \text{ mol}
\][/tex]
So, the number of moles of pure KOH is approximately 1.426 mol.
### 7.5.3
Next, determine which reactant is in excess and state whether the final solution is acidic, basic, or neutral.
1. Calculate the moles of H₂SO₄:
Given that we have 100 mL of a 1 M H₂SO₄ solution:
[tex]\[
\text{Moles of H₂SO₄} = \text{Volume (L)} \times \text{Concentration (M)} = 0.100 \text{ L} \times 1 \text{ M} = 0.1 \text{ mol}
\][/tex]
2. Determine the stoichiometry of the reaction:
The balanced chemical equation for the reaction between KOH and H₂SO₄ is:
[tex]\[
2 \text{ KOH} + \text{H}_2\text{SO}_4 \rightarrow \text{K}_2\text{SO}_4 + 2 \text{H}_2\text{O}
\][/tex]
According to the equation, 2 moles of KOH react with 1 mole of H₂SO₄.
3. Calculate the required moles of KOH:
[tex]\[
\text{Required moles of KOH} = 2 \times \text{Moles of H₂SO₄} = 2 \times 0.1 \text{ mol} = 0.2 \text{ mol}
\][/tex]
4. Compare the moles of available KOH with the required moles:
[tex]\[
\text{Available moles of KOH} = 1.426 \text{ mol}
\][/tex]
[tex]\[
\text{Required moles of KOH} = 0.2 \text{ mol}
\][/tex]
Since the available moles of KOH (1.426 mol) are much greater than the required moles (0.2 mol), KOH is the excess reactant.
5. Determine the nature of the final solution:
As KOH is in excess, the solution will be basic.
So, the final solution is BASIC.
