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Q.

28 g of N2 and 6 g of H2 were mixed. At equilibrium 17 g NH3 was produced. The weight of N2 and H2 at equilibrium are respectively

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a

11 g, 0 g

b

1 g, 3 g

c

14 g, 3 g

d

11 g, 3 g

answer is C.

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Detailed Solution

N2      +         3H2       →       2NH3t = 0Initial       2828               62                      0                 1                    3                       0                mole             mole                 moleAt equilibriumt = ?           1 - x            3 - 3x                  2x                  2x=1717     2x = 1               x = 1/2Final mole     1/2            3/2                       1weight = 14 g  3 g
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