When 0.087 mol of NO and 0.0437 mol of $B r_{2}$ are mixed in a closed container at constant temperature, 0.0518 mol of NOBr is obtained at equilibrium. The equilibrium amount of NO and $B r_{2}$ are:

N O = 0.352 m o l, B r_{2} = 0.0178 m o l

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N O = 0.0352 m o l, B r_{2} = 0.178 m o l

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N O = 0.352 m o l, B r_{2} = 0.178 m o l

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N O = 0.0352 m o l, B r_{2} = 0.0178 m o l

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Solution

The correct option is D $N O = 0.0352 m o l, B r_{2} = 0.0178 m o l$
The equilibrium reaction is as shown below.
$2 N O (g) + B r_{2} (g) ⇌ 2 N O B r (g)$

NO
$B r_{2}$
NOBr
Initial
0.087
0.043
0
Change
-2x
-x
2x
Equilibrium
0.087-2x
0.043-x
2x
The number of moles of NOBr at equilibrium are 0.0518.
Thus $2 x = 0.0518 m o l$ or $x = 0.0259 m o l$ .
The number of moles of $N O = (0.087 - 0.0518) m o l = 0.0352 m o l$ .
The number of moles of $B r_{2} = (0.043 - 0.0259) m o l = 0.0178 m o l$ .

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