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The equilibrium constant of the reaction...

The equilibrium constant of the reaction `A_(2)(g)+B_(2)(g)hArr 2AB(g)` at 373 K is 50. If 1 L of flask containing 1 mole of `A_(2)(g)` is connected to 2L flask containing 2 moles `B_(2)(g)` at `100^(@)C`, the amount of AB produced at equilibrium at `100^(@)C` would be

A

0.93 mol

B

1.87 mol

C

2.80 mol

D

3.74 mol

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The correct Answer is:
B
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The equilibrium constant of the reaction A_(2)(g)+B_(2)(g) hArr 2AB(g) at 100^(@)C is 50 . If a 1-L flask containing 1 mol of A_(2) is connected to a 2 L flask containing 2 mol of B_(2) , how many moles of AB will be formed at 373 K ?

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