How many millilitres of `H_(2)`O vapour measured at `327^(@)`C and 760 torr are formed when 50 mL of ammonia at 950 torr and `127^(@)`C reacts with oxygen accordirig to the following reaction?
`NH_(3) (g) + O_(2) (g) to N_(2) (g) + H_(2) O(g)`

Options

A. 76 mL
B. 125 mL
C. 140.625 mL
D. 241.4 mL

Answer

C

Explanation

We first require volume of ammonia at the same condition as it is for water vapour . Hence , we use combined gas law equation `(p_(2) V_(2))/(T_(2)) = (p_(1) V_(1))/(T_(1)) implies V_(2) = ((p_(1))/(p_(2))) ((T_(2))/(T_(1))) V_(1) = (950)/(760) xx (600)/(400) xx 50 ` mL = 93.75 mL
Now considering the following balanced chemical equation , we get `4H_(3) (g) + 3O_(2) (g) to2 N_(2) (g) + 6 H_(2) O(g)`
4 vol of `NH_(3) = 6` vol of `H_(2)O implies 93.75` mL of `NH_(3)` will produce = `(6)/(4) xx 93.75` mL = 140.625 mL of `H_(2)O` vapour