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50.0 kg of N(2) (g) and 10.0 kg of H(2) ...

50.0 kg of `N_(2) (g)` and 10.0 kg of `H_(2)` (g) are mixed, to produce `NH_(3)(g)`. Calculate the `NH_(3)(g)`. Formed. Identify the limiting reagent in the production of `NH_(3)` in this situation.

Text Solution

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According to the balanced chemical reaction, `25 kg` of `N_(2)` reacts with `6 kg` of `H_(2)` to give `34 kg` of `NH_(3)`.
Thus, `6 kg` of `H_(2)` requires `-= 25 kg of N_(2)`
`10 kg` of `H_(2)` requries `-= (28 xx 10)/(6) = 46.6 kg or N_(2)`
Hence, `H_(2)` is the limiting reagent, since it is completely consumed in the reaction, and `H_(2)` will decied the amount of product formd.
Thus, `6 kg of H_(2) -= 34 kg of NH_(3)` ltbgt `10 kg of H_(2) -= (34 xx 10)/(6) = 56.6 kg of NH_(3)`
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Nitrogen and hydrogen react to form ammonia according to the reaction: N_(2)(g)+3H_(2)(g)to2NH_(3)(g) If 1000 g H_(2) react with 2000g of N_(2) (i) Identify the limiting reagent. (ii) Calculate the mass of ammonia (NH_(3)) which will be formed.

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Knowledge Check

  • 20.0 kg of N_2(g) and 3.0 kg of H_2(g) are mixed to produce NH_3(g) . The amount of NH_3 (g) formed is

    A
    17 g
    B
    34 g
    C
    20 g
    D
    3 kg

  • For a reaction, N_2(g) +3H_2(g)rarr2NH_3(g) Identify dihydrogen (H_2) as a limiting reagent in the following reaction mixtures.

    A
    `35 g of N_2 + 8 g of H_2`
    B
    `28 g of N_2 + 6 g of H_2`
    C
    `56 g of N_2 + 12 g of H_2`
    D
    `14 g of N_2 + 4 g of H_2`

  • For the process NH_3(g)+HCI(g) rarr NH_4CI(s)

    A
    Both `DeltaH` and `DeltaS` are + ve
    B
    `DeltaH` is -ve and `DeltaS` is + ve
    C
    `DeltaH` is + ve and `DeltaS` is -ve
    D
    Both `DeltaH` and `DeltaS` are -ve

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