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Answer :
The amount of sulfuric acid produced when 68.1 g of lead(II) sulfate react is 32.3 g. None of the given options is answer.
To calculate the amount of sulfuric acid produced, we need to use the balanced chemical equation for the reaction that occurs when an automobile battery is charged.
Given:
- 68.1 g of lead(II) sulfate (PbSO[tex]_4[/tex]) react
Step 1: Write the balanced chemical equation for the reaction:
2PbSO[tex]_4[/tex] + 2H[tex]_2[/tex]O ⇌ 2PbO[tex]_2[/tex] + 2H[tex]_2[/tex]SO[tex]_4[/tex]
Step 2: Calculate the moles of lead(II) sulfate:
- Moles of PbSO[tex]_4[/tex] = 68.1 g / (207.2 g/mol)
- Moles of PbSO = 0.329 mol
Step 3: Use the balanced equation to calculate the moles of sulfuric acid (H2SO[tex]_4[/tex]) produced:
2 mol H[tex]_2[/tex]SO[tex]_4[/tex]are produced for every 2 mol PbSO[tex]_4[/tex] that react
- Moles of H[tex]_2[/tex]SO[tex]_4[/tex] produced = 0.329 mol PbSO[tex]_4[/tex] * (2 mol H[tex]_2[/tex]SO[tex]_4[/tex] / 2 mol PbSO[tex]_4[/tex])
- Moles of H[tex]_2[/tex]SO[tex]_4[/tex] produced = 0.329 mol
Step 4: Calculate the mass of sulfuric acid produced:
- Mass of H[tex]_2[/tex]SO[tex]_4[/tex] = 0.329 mol * (98.08 g/mol)
- Mass of H[tex]_2[/tex] SO[tex]_4[/tex] = 32.3 g
Therefore, the amount of sulfuric acid produced when 68.1 g of lead(II) sulfate react is 32.3 g.
None of the given options is answer.
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