`50.0 kg` of `N_(2) (g)` and `10 kg` of `H_(2) (g)` are mixed to produce `NH_(3) (g)`. Calculate the `NH_(3) (g)` formed. Identify the limiting reagent.

Explanation

A balanced equation for the above reaction is written as follows :
`N_(2) ` (g) `3H_(2) (g) hArr 2 NH_(3) (g) `
Calculation of moles :
Number of moles of `N_(2)`
`=50.0 kg N_(2) xx (1000 g N_(2) )/(1 kg N_(2)) xx ( 1 mol M_(2) )/( 2.016 g H_(2))`
`=4.96 xx10^(3)` mol
According to the above equation, 1 mol `N_(2)` (g) requires 3 mol` H_(2)` (g), for the reaction. Hence, for `17.86xx102` mol of` N_(2)`, the moles of `H_(2)` (g) required would be
`=17.86 xx10^(3) Mol N_(2) xx( 3 mol H_(2) (g) )/(1 mol N_(2) (g) )`
`=5.36xx10^(3)` mol `H_(2) `
But we have only `4.96xx10^(3) `mol `H_(2)`. Hence, dihydrogen is the limiting reagent in this case. So, `NH_(3)`(g) would be formed only from that amount of available dihydrogen i.e.,` 4.96 xx 10^(3) `mol
Since 3 mol `H_(2) (g) ` gives 2 mol `NH_(3) (g) `
` 4.96xx10^(3) mol H_(2) (g) xx(2 mol NH_(3) (g) )/( 3 mol H_(2) (g) )`
`=3.30 xx10^(3) `mol `NH_(3)` (g) is obtained .
If they are to be converted to grams, it is done as follows :
1 mol `NH_(3) (g) = 17.0 g NH _(3) (g) `
` 3.30xx10^(3) ` mol `NH_(3) (g) xx ( 17.0 g NH_(3) (g))/( 1 mol NH_(3) (g) )`
`= 3.30 xx10^(3) xx 17 g NH_(3) (g) `
` = 56.1 xx 10 ^(3) g NH_(3) `
` =56.1 kg NH _(3) `