Question

What volume of oxygen (in L) is produced

Answer 1

Answer:

12.36 L.

Explanation:

We'll begin by calculating the number of mole in 147.1 g of lead(II) nitrate, Pb(NO₃)₂. This can be obtained as follow:

Molar mass of Pb(NO₃)₂ = 207.2 + 2[14.01 + (16×3)]

= 207.2 + 2[14.01 + 48]

= 207.2 + 2[62.01]

= 207.2 + 124.02

= 331.22 g/mol

Mass of Pb(NO₃)₂ = 147.1 g

Mole of Pb(NO₃)₂ =?

Mole = mass / Molar mass

Mole of Pb(NO₃)₂ = 147.1 / 331.22

Mole of Pb(NO₃)₂ = 1.104 moles.

Next, we shall determine the number of mole of oxygen gas, O₂, produce from the reaction. This can be obtained as follow:

2Pb(NO₃)₂ —> 2PbO + 4NO₂ + O₂

From the balanced equation above,

2 moles of Pb(NO₃)₂ decomposed to produce 1 mole of O₂.

Therefore, 1.104 moles of Pb(NO₃)₂ will decompose to produce = (1.104 × 1)/2 = 0.552 mole of O₂.

Finally, we shall determine the volume occupied by 0.552 mole of oxygen gas, O₂. This can be obtained as follow:

1 mole of O₂ occupied 22.4 L at STP.

Therefore, 0.552 mole of O₂ will occupy = 0.552 × 22.4 = 12.36 L at STP.

Thus, the volume of oxygen gas, O₂ produced is 12.36 L.