2NH3(g)+3CuO(s)= + 3H2O(I)+N2(g) If a sample of 51.0g of ammonia’s reacted with excess copper oxide, how many moles of nitrogen would

Question

2NH3(g)+3CuO(s)= + 3H2O(I)+N2(g)
If a sample of 51.0g of ammonia’s reacted with excess copper oxide, how many moles of nitrogen would be produced?

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Hải Đăng 4 years 2021-09-02T06:18:53+00:00 1 Answers 192 views 0

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  1. diemkieu
    0
    2021-09-02T06:20:40+00:00
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    Answer:

    1.50 mol N₂

    Explanation:

    Step 1: Write the balanced equation

    2 NH₃(g) + 3 CuO(s) ⇒ 3 Cu + 3 H₂O(I) + N₂(g)

    Step 2: Calculate the moles corresponding to 51.0 g of NH₃

    The molar mass of NH₃ is 17.03 g/mol.

    51.0 g × 1 mol/17.03 g = 2.99 mol

    Step 3: Calculate the moles of N₂ produced from 2.99 moles of NH₃

    The molar ratio of  NH₃ to N₂is 2:1. The moles of N₂ produced are:

    2.99 mol NH₃ × 1 mol N₂/2 mol NH₃ = 1.50 mol N₂

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