To solve the problem step by step, we need to determine the new percentage composition of the oleum sample after adding 0.45 g of water (H₂O).
### Step 1: Calculate the moles of water added
To find the moles of water added, we use the formula:
\[
\text{Moles of H}_2\text{O} = \frac{\text{Given mass}}{\text{Molar mass}}
\]
The molar mass of water (H₂O) is approximately 18 g/mol.
Given mass = 0.45 g
\[
\text{Moles of H}_2\text{O} = \frac{0.45 \, \text{g}}{18 \, \text{g/mol}} = 0.025 \, \text{mol}
\]
### Step 2: Determine the moles of SO₃ in the oleum sample
Next, we need to find the moles of SO₃ in the oleum sample. The molar mass of SO₃ is approximately 80 g/mol.
Given mass of SO₃ = 10 g
\[
\text{Moles of SO}_3 = \frac{10 \, \text{g}}{80 \, \text{g/mol}} = 0.125 \, \text{mol}
\]
### Step 3: Determine the reaction between SO₃ and H₂O
In the reaction between SO₃ and H₂O, the following reaction occurs:
\[
\text{SO}_3 + \text{H}_2\text{O} \rightarrow \text{H}_2\text{SO}_4
\]
From the stoichiometry of the reaction, 1 mole of SO₃ reacts with 1 mole of H₂O. Therefore, the moles of SO₃ that will react with the added water is equal to the moles of water added, which is 0.025 mol.
### Step 4: Calculate the mass of SO₃ that reacts
To find the mass of SO₃ that reacts, we use the moles of SO₃ that react and its molar mass:
\[
\text{Mass of SO}_3 \text{ that reacts} = \text{Moles of SO}_3 \times \text{Molar mass of SO}_3
\]
\[
\text{Mass of SO}_3 \text{ that reacts} = 0.025 \, \text{mol} \times 80 \, \text{g/mol} = 2 \, \text{g}
\]
### Step 5: Calculate the remaining mass of SO₃
Now, we can find the remaining mass of SO₃ after the reaction:
\[
\text{Remaining mass of SO}_3 = \text{Initial mass} - \text{Mass that reacted}
\]
\[
\text{Remaining mass of SO}_3 = 10 \, \text{g} - 2 \, \text{g} = 8 \, \text{g}
\]
### Step 6: Calculate the total mass of the oleum sample after adding H₂O
The total mass of the oleum sample after adding 0.45 g of water is:
\[
\text{Total mass} = \text{Mass of SO}_3 + \text{Mass of H}_2\text{SO}_4 + \text{Mass of added H}_2\text{O}
\]
Given mass of H₂SO₄ = 15 g
\[
\text{Total mass} = 8 \, \text{g (remaining SO}_3) + 15 \, \text{g (H}_2\text{SO}_4) + 0.45 \, \text{g (H}_2\text{O}) = 23.45 \, \text{g}
\]
### Step 7: Calculate the new percentage of SO₃ in the oleum sample
Finally, we can calculate the new percentage of SO₃ in the oleum sample:
\[
\text{Percentage of SO}_3 = \left( \frac{\text{Remaining mass of SO}_3}{\text{Total mass}} \right) \times 100
\]
\[
\text{Percentage of SO}_3 = \left( \frac{8 \, \text{g}}{23.45 \, \text{g}} \right) \times 100 \approx 34.1\%
\]
### Final Answer
The new percentage of SO₃ in the oleum sample after adding 0.45 g of water is approximately **34.1%**.
