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. 51 kg
C. 60 kg
D. 34 kg

Answer

A

Explanation

`N_(2) + 3H_(2) to 2NH_(3)`
1 mol of `N_(2)` (28 g) combine with 3 mol of (6g) of `H_(2)`
`therefore 3 kg` of `N_(2)` will react with `28/3 xx 3 = 14 kg` of `N_(2)`
`H_(2)` is the limiting reagent.
6g of `H_(2)` produce `NH_(3) = 34 g`
3.0 kg of `H_(2)` will produce `NH_(3) = 34/6 xx 3 = 17`kg