using the balanced equation below how many grams of lead(||) sulfate would be produced from the complete reaction of 23.6 g lead (|V) oxide
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Answer:
59.8 g of PbSO₄.
Explanation:
The balanced equation for the reaction is given below:
Pb + PbO₂ + 2H₂SO₄ —> 2PbSO₄ + 2H₂O
Next, we shall determine the mass of PbO₂ that reacted and the mass of PbSO₄ produced from the balanced equation. This can be obtained as follow:
Molar mass of PbO₂ = 207 + (16×2)
= 207 + 32
= 239 g/mol
Mass of PbO₂ from the balanced equation = 1 × 239 = 239 g
Molar mass of PbSO₄ = 207 + 32 + (16×4)
= 207 + 32 + 64
= 303 g/mol
Mass of PbSO₄ from the balanced equation = 2 × 303 = 606 g
SUMMARY:
From the balanced equation above,
239 g of PbO₂ reacted to produce 606 g of PbSO₄.
Finally, we shall determine the mass of PbSO₄ that will be produced by the reaction of 23.6 g of PbO₂. This can be obtained as follow:
From the balanced equation above,
239 g of PbO₂ reacted to produce 606 g of PbSO₄.
Therefore, 23.6 g of PbO₂ will react to produce = (23.6 × 606) / 239 = 59.8 g of PbSO₄.
Thus, 59.8 g of PbSO₄ were obtained from the reaction.