Nitric onxide reacts with bromine to from nitrosyl bromide as follows 2NO(g)+Br_2(g)iff2NOBr(g) When 0.087 mol of NO and 0.0437 mole of Br_2 are mixed in a closed containar at constant temperature,0.0518 mol of NOBr is obtained at equilibrium. Calculate equilibrium amount of NO and Br_2.

Nitric onxide reacts with bromine to from nitrosyl bromide as follows

1.	Nitric onxide reacts with bromine to from nitrosyl bromide as follows 2NO(g)+Br_2(g)iff2NOBr(g) When 0.087 mol of NO and 0.0437 mole of Br_2 are mixed in a closed containar at constant temperature,0.0518 mol of NOBr is obtained at equilibrium. Calculate equilibrium amount of NO and Br_2.
Answer» Solution :2 moles of NO gives 2 MOLE of NOBR `therefore` Number of moles NO REACTED to from 0.0518 MOL of NOBr=0.0518 Number of moles of `Br_2` reacted= `0.0518/2=0.0259` `therefore` Number of moles of NO remaining at equilibrium=0.087-0.0518=0.0352 mol `therefore` Number of moles `Br_2` at equilibrium =0.0437-0.0259=0.0178 mol