How many mililitres 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 according to the following reaction?
`NH_(3)(g) +O_(2)(g) to N_(2)(g)+H_(2)O(g)`

Options

A. `75 mL`
B. `125 mL`
C. `140.22 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 50mL= 93.75mL`
Now considering the following balanced chemical equation, we get `4H_(3)(g)+3O_(2)(g) to 2N_(2)(g)+6H_(2)O(g)`
`4" vol of "NH_(3)= 6" vol of "H_(2)O implies 93.75mL" of "NH_(3)` will produce
`=(6)/(4)xx 93.75mL= 140.625mL H_(2)O` vapour.