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

Explanation

(a) (1) `underset(2 "moles")(2NH_(3(g))+underset(1" mole "CO_(2)(g))tounderset(1" mole") ((NH_(4))_(2)CO_((aq)))+H_(2)O_(1)`
No. of moles of ammonia -=`(637.2)/(17)`=37.45 mol e
No. of moles of `CO_(2)=(1142)/(44)` = 25.95 moles
As per the balanced equation, one mole of Co, requires 2 moles of ammonia. No. of moles of NH, required to react with 25.95 moles of `CO_(2)`, is =`(2)/(1)xx25.95`=51.90moles
`therefore`   37.45 moles of NH is not enough to completely react with `CO_(2)` (25.95 moles). Hence, NH must be the limiting reagent, and`CO_(2)`, is excess reagent.
(ii) 2 moles of ammonia produce mole of urea. `therefore` Limiting reagent 37.45 moles of `NH_(3)`, can produce `(1)/(2)xx37.45` moles of urea. = 18.725 moles of urca.
The mass of 18.725 moles of urca =No. of moles` xx`Molar mass
= `18.725xx60` =1123.5 g of urea.
2 moles of ammonia requires mole of `CO_(2)`
`therefore`   Limiting reagent 37.45 mol es of NII, will require `(1)/(2)xx37.45mol es of CO_(2)xx37.45` moles of `CO_(2)` =18.725 moles of `CO_(2)`
  No. of moles of the excess reagent (CO) lell -2595-18725 - =7225
The mass of the excess reagent `(CO_(2))`left =`7225xx44` = 3179g of `CO_(2)`