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

`N_2+3H_2 hArr 2NH_3`
`28 (g). 6g 34(g)`
`6g H_2` reacts with `28g N_2`
`10g H_2` reacts with `(28)/(6)xx10=46.67g N_2`.
`H_2` is limiting reagent where as `N_2` is excess.
`6g H_2` produces `34 Gnh_3`
`10g H_2 produces ((34)/(6)xx10)g NH_3`
To calculate number of moles of `NH_3`
`n NH_3 =("Product weight")/("Molecular weight")=((34)/(6)xx10))/(17)=(34)/(6)xx(10)/(17)=(10)/(3)=3.33`.