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11 March, 12:49

How many grams of N2 are required to produce 100.0 L of NH3 at STP?

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Answers (2)
  1. 11 March, 16:10
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    There is no equation given so I'll assume it's just the basic equation: N2 + 3H2 - > 2NH3

    Molar mass of N2 = 28g

    Since the two data sets are not in the same unit we first need to find the moles of gas from the given volume so we use: PV = nRT

    where: P = 1 atm, V = 100ltr, T = 273 K

    n = (1 atm) (100 L) / (0.08206 L-atm/mol-K) (273K) n = 4.46 mol of NH3

    from the reaction equation we can see that 1 mole of N2 yields 2 moles of NH3 therefore:

    4.46 mol NH3 (1 mol N2/2 mol NH3) = 2.23 moles N2

    2.23 moles of N2 (28 g N2/1 mol N2) = 62.5g N2
  2. 11 March, 16:45
    0
    Given:

    N2, volume of 1L of NH3 at STP

    Required:

    Grams of N2

    Solution:

    N2 + 3H2 - > 2NH3

    Molar mass of N2 = 28g

    From the ideal gas equation PV = nRT

    n = PV/RT

    n = (1 atm) (100 L) / (0.08206 L-atm/mol-K) (273K)

    n = 4.46 mol of NH3

    from the reaction, we need 2 moles of NH3 to get 1 mole of N2

    4.46 mol NH3 (1 mol N2/2 mol NH3) = 2.23 moles N2

    2.23 moles of N2 (28 g N2/1 mol N2) = 62.5g N2
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