20.0 kg of `N_(2)(g)` and 3.0 kg of `H_(2)(g)` are mixed to produce `NH_(3)(g)`. The amount of `NH_(3)(g)` formed is

Options

A. 17 kg
B. 34 kg
C. 20 kg
D. 3 kg

Answer

A

Explanation

`underset(28" g")(N_(2)) + underset(6" g")(3H_(2)) rarr underset(34" g")(2NH_(3))` Limiting reactant is `H_(2)`
Weight of `NH_(3)` obtained from `6xx 10^(-3) ` kg of `H_(2)`
`= 34xx 10^(-3) ` kg
`therefore` Weight of `NH_(3)` obtained from 3 kh of `H_(2)`
`= (34xx 10^(-3))/(6xx 10^(-3)) xx 3 `
= 17 kg `NH_(3)`