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1 November, 23:12

If 3.52 L of nitrogen gas and 2.75 L of hydrogen gas were allowed to react, how many litres of ammonia gas could form? Assume all gases are at the same temperature and pressure.

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  1. 2 November, 03:01
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    V NH3 (g) = 1.833 L

    Explanation:

    balanced reaction:

    N2 (g) + 3H2 (g) → 2NH3 (g)

    assuming STP:

    ∴ V N2 (g) = 3.52 L

    ∴ V H2 (g) = 2.75 L

    ideal gas:

    PV = RTn

    ∴ moles N2 (g) = PV/RT

    ⇒ mol N2 (g) = (1 atm) (3.52 L) / (0.082 atm. L/K. mol) (298 K)

    ⇒ mol N2 (g) = 0.144 mol

    ∴ moles H2 (g) = PV/RT

    ⇒ mol H2 (g) = (1) (2.75) / (0.082) (298) = 0.113 mol (limit reagent)

    ∴ moles NH3 (g) = (0.113 moles H2 (g)) (2 moles NH3 / 3 mol H2) = 0.075 mol

    ∴ V NH3 (g) = RTn/P

    ⇒ V NH3 (g) = ((0.082 atm. L/K. mol) (298 K) (0.075 mol)) / (1 atm)

    ⇒ V NH3 (g) = 1.833 L
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