In Haber's process 50.0g of `N_(2)` [g] and 10.0 g of `H_(2)` [g] are mixed to produce `NH_(3)` [g]. What is the number of moles of `NH_(3)` [g] formed?

Options

A. 3.33
B. 2.36
C. 2.01
D. 5.36

Answer

a

Explanation

The reaction proceeds as :
`N_(2) + 3H_(2) hArr 2NH_(3)`
`28 g 6 g 34 g`
`:' 6g H_(2)` reacts with `28 g N_(2)`
`:. 10 g H_(2)` reacts with `(28)/(6) xx 10 = 46.67 g N_(2)`
i.e., `H_(2)` is limiting reagent and `N_(2)` is in excess.
Now
`:' 6g H_(2)` produced `34 g NH_(3)`
`:. 10 g H_(2)` produces `((34)/(6) xx 10) g NH_(3))`
Therefore, we can calculate number of moles of `NH_(3)` as `.^(n)NH_(3) = ("produced weight")/("Molecular weigth")`
`(((34)/(6) xx 10))/(17) = (34)/(6) xx (10)/(17) = (10)/(3) = 3.33`