Urea is prepared by the reaction between ammonia and carbon dioxide.
`2NH_(3_((g)))+CO_(2_((g)))rarr(NH_(4))_(2)CO_(aq)+H_(2)O_((l))`
In one process, 637.2 g of `NH_(3)` are allowed to react with 1142 g of `CO_(2)`
(a) Which of the two reactants is the limiting reagent?
`2NH_(3_((g)))+CO_(2_((g)))rarr(NH_(4))_(2)CO_(aq)+H_(2)O_((l))`
(b) Calculate the mass of `(NH_(4))_(2)CO` formed.
`2NH_(3_((g)))+CO_(2_((g)))rarr(NH_(4))_(2)CO_(aq)+H_(2)O_((l))`
(c) How much of the excess reagent in grams is left at the end of the reaction?

Explanation

(a) `2NH_(3_((g))) + CO_(2((g))) rarr (NH_(4))_(2)CO_(aq) + H_(2)O_((l))`
No. of moles of ammonia =`637.3/17` = 37.45 mole
No. of moles of `CO_(2)` =`1142/44`=25.95 moles
As per the balanced equation, one mole of `CO_(2)` requires 2 moles of ammonia.
`:.` No. of moles of `NH_(3)` required to react with 25.95 moles of `CO_(2)` is =
`2/1 xx 25.95` = 51.90 moles.
`:.` 37.45 moles of `NH_(3)` is not enough to completely react with `CO_(2)` (25.95 moles).
Hence, `NH_(3)` must be the limiting reagent, and `CO_(2)` is excess reagent.
(b) 2 moles of ammonia produce 1 mole of urea.
`:.` Limiting reagent 37.45 moles of `NH_(3)` can produce `1/2 xx 37.45`moles of urea
= 18.725 moles of urea.
The mass of 18.725 moles of urea = No. of moles `xx` Molar mass
= 18.725 `xx` 60 = 1123.5 g of urea.
(c) 2 moles of ammonia requires 1 mole of `CO_(2)`
`:.` Limiting reagent 37.45 moles of `NH_(3)` will require` 1/2 xx 37.45` moles of `CO_(2)`
= 18.725 moles of `CO_(2)`
`:.` No. of moles of the excess reagent `(CO_(2))` left = 25.95- 18.725 — 7.225
The mass of the excess reagent `(CO_(2))` left = `7.225xx44`
= 317.9 g of `CO_(2)` .