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

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.

frequency, z, and the collision density, z, in carbon dioxide (co2), r 180 pm at 30 °c and 120 kpa

pv=nrt

explanation: