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If 800 cm3 of H2 gas reacts with N2 gas to form NH3 gas according to the equation 3H2 + N2 -> 2NH3, what volume of N2 is used and what volume of NH3 is produced?

Chemistry

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Answer

2=266.67 cm3_2 = 266.67 \text{ cm}^3

3=533.33 cm3_3 = 533.33 \text{ cm}^3

Step-by-step solution

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The balanced chemical equation is:

3H2(g)+N2(g)⟶2NH3(g)3\text{H}_2(\text{g}) + \text{N}_2(\text{g}) \longrightarrow 2\text{NH}_3(\text{g})

According to Gay-Lussac's Law of Combining Volumes, the ratio of the volumes of reacting gases and gaseous products is given by their stoichiometric coefficients, assuming constant temperature and pressure.

  1. Determine the volume of Nitrogen used.

    From the equation, 3 volumes of H2_2 react with 1 volume of N2_2.

    Volume of N2Volume of H2=13\frac{\text{Volume of N}_2}{\text{Volume of H}_2} = \frac{1}{3} Volume of N2=13×800 cm3\text{Volume of N}_2 = \frac{1}{3} \times 800 \text{ cm}^3 Volume of N2=266.67 cm3\text{Volume of N}_2 = \boxed{266.67 \text{ cm}^3}

  2. Determine the volume of Ammonia produced.

    From the equation, 3 volumes of H2_2 produce 2 volumes of NH3_3.

    Volume of NH3Volume of H2=23\frac{\text{Volume of NH}_3}{\text{Volume of H}_2} = \frac{2}{3} Volume of NH3=23×800 cm3\text{Volume of NH}_3 = \frac{2}{3} \times 800 \text{ cm}^3 Volume of NH3=533.33 cm3\text{Volume of NH}_3 = \boxed{533.33 \text{ cm}^3}
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