The enthalpy of the reaction forming PbO according to the following equation is 438 kJ. What heat energy (kJ) is releated in formation of 22.3 g PbO(s)?
(Atomic masses : Pb = 207, O = 16.0)
`2Pb(s) +O_(2)(g)rarr2PbO(s)`

Options

A. 21.9
B. 28.7
C. 14.6
D. 34.2

Explanation

To solve the problem, we need to determine the heat energy released during the formation of 22.3 g of PbO(s) given that the enthalpy of the reaction forming PbO is 438 kJ for the reaction:

\[ 2 \text{Pb}(s) + \text{O}_2(g) \rightarrow 2 \text{PbO}(s) \]

### Step-by-Step Solution:

**Step 1: Determine the molar mass of PbO.**

The molar mass of PbO can be calculated by adding the atomic masses of lead (Pb) and oxygen (O):
- Atomic mass of Pb = 207 g/mol
- Atomic mass of O = 16 g/mol

\[ \text{Molar mass of PbO} = 207 + 16 = 223 \text{ g/mol} \]

**Step 2: Calculate the number of moles of PbO in 22.3 g.**

Using the formula for moles:

\[ \text{Number of moles} = \frac{\text{mass (g)}}{\text{molar mass (g/mol)}} \]

Substituting the values:

\[ \text{Number of moles of PbO} = \frac{22.3 \text{ g}}{223 \text{ g/mol}} \approx 0.1 \text{ moles} \]

**Step 3: Calculate the enthalpy change for the formation of 1 mole of PbO.**

The enthalpy change given is for the formation of 2 moles of PbO, which is 438 kJ. Therefore, the enthalpy change for the formation of 1 mole of PbO is:

\[ \text{Enthalpy change for 1 mole of PbO} = \frac{438 \text{ kJ}}{2} = 219 \text{ kJ} \]

**Step 4: Calculate the heat energy released for 0.1 moles of PbO.**

Now, we can find the heat energy released for 0.1 moles of PbO:

\[ \text{Heat released} = \text{Number of moles} \times \text{Enthalpy change for 1 mole} \]

Substituting the values:

\[ \text{Heat released} = 0.1 \text{ moles} \times 219 \text{ kJ/mol} = 21.9 \text{ kJ} \]

### Final Answer:

The heat energy released in the formation of 22.3 g of PbO(s) is **21.9 kJ**.

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