To solve the problem, we first need to understand the reaction between lead oxide (PbO) and ammonia (NH₃). The reaction can be represented by the following balanced equation:
\[ 3 \text{PbO} + 2 \text{NH}_3 \rightarrow 3 \text{Pb} + \text{N}_2 + 3 \text{H}_2\text{O} \]
From the equation:
- 3 moles of PbO yield 3 moles of Pb, 1 mole of N₂, and 3 moles of H₂O.
Now, let's calculate the molar masses of the relevant substances:
- Molar mass of PbO = 207.2 (Pb) + 16.0 (O) = 223.2 g/mol
- Molar mass of NH₃ = 14.0 (N) + 3 × 1.0 (H) = 17.0 g/mol
- Molar mass of Pb = 207.2 g/mol
- Molar mass of N₂ = 28.0 g/mol (since it is diatomic)
- Molar mass of H₂O = 18.0 g/mol
Now, we can calculate the number of moles of each reactant:
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Moles of PbO: \[ \text{Moles of PbO} = \frac{30 \text{ g}}{223.2 \text{ g/mol}} \approx 0.134 \text{ moles} \]
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Moles of NH₃: \[ \text{Moles of NH₃} = \frac{15 \text{ g}}{17.0 \text{ g/mol}} \approx 0.882 \text{ moles} \]
Next, let's determine the limiting reactant:
From the balanced equation:
- 3 moles of PbO require 2 moles of NH₃. Therefore, 0.134 moles of PbO would require: \[ \text{Required moles of NH₃} = 0.134 \text{ moles} \times \frac{2}{3} \approx 0.089 \text{ moles} \]
Since we have approximately 0.882 moles of NH₃ and only need about 0.089 moles, PbO is the limiting reactant.
Now, let's find the mass of the products formed:
From the balanced equation:
- From 3 moles of PbO, we produce 3 moles of Pb, 1 mole of N₂, and 3 moles of H₂O.
Using the moles of the limiting reactant (PbO), we can calculate the moles of each product:
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Moles of Pb produced from 0.134 moles of PbO: \[ \text{Moles of Pb} = 0.134 \text{ moles} \times \frac{3}{3} = 0.134 \text{ moles} \] Mass of Pb: \[ \text{Mass of Pb} = 0.134 \text{ moles} \times 207.2 \text{ g/mol} \approx 27.8 \text{ g} \]
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Moles of N₂ produced: \[ \text{Moles of N₂} = 0.134 \text{ moles} \times \frac{1}{3} \approx 0.0447 \text{ moles} \] Mass of N₂: \[ \text{Mass of N₂} = 0.0447 \text{ moles} \times 28.0 \text{ g/mol} \approx 1.25 \text{ g} \]
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Moles of H₂O produced: \[ \text{Moles of H₂O} = 0.134 \text{ moles} \times \frac{3}{3} = 0.134 \text{ moles} \] Mass of H₂O: \[ \text{Mass of H₂O} = 0.134 \text{ moles} \times 18.0 \text{ g/mol} \approx 2.41 \text{ g} \]
Now, let's sum the masses of the products: \[ \text{Total mass} = \text{mass of Pb} + \text{mass of N₂} + \text{mass of H₂O} \approx 27.8 \text{ g} + 1.25 \text{ g} + 2.41 \text{ g} \approx 31.46 \text{ g} \]
The approximate mass of all three products (Pb, N₂, and H₂O) is around 31.5 grams.
Based on the provided options (10 g, 15 g, 45 g, 450 g), the closest correct answer would be 45 g.