Thirty grams of lead oxide and fifteen grams of ammonia react completely to produce solid lead, nitrogen gas, and liquid water. What will be approximate mass of all three products?

10 g
15 g
45 g
450 g

1 answer

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:

  1. Moles of PbO: \[ \text{Moles of PbO} = \frac{30 \text{ g}}{223.2 \text{ g/mol}} \approx 0.134 \text{ moles} \]

  2. 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:

  1. 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: \[ \text{Mass of Pb} = 0.134 \text{ moles} \times 207.2 \text{ g/mol} \approx 27.8 \text{ g} \]

  2. Moles of N₂ produced: \[ \text{Moles of N₂} = 0.134 \text{ moles} \times \frac{1}{3} \approx 0.0447 \text{ moles} \] Mass of N₂: \[ \text{Mass of N₂} = 0.0447 \text{ moles} \times 28.0 \text{ g/mol} \approx 1.25 \text{ g} \]

  3. 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.