Question

Limiting reactant: Urea `[(NH_(2))_(2) CO]` used as ferlilzer as animal feed, and in polymer industry, is prepared by the reaction between ammonia and carbon dioxide:
`2NH_(3)(g) + CO_(2) (g) rarr (NH_(2))_(2) CO(aq.) + H_(2) O(1)`
In one process , `637.2g` of `NH_(3)` is allowed to react with `11.42g` of `CO_(2)`
(i) Which of the two reactants is the limiting reactant?
(ii) Calculate the mass of `(NH_(2))_(2) CO` formed?
(iii) How much of the excess reagent (in grams) is left at the end of the reaction?
Strategy: (i) Since we cannot tell by inspection which of the two recantants is the limiting reacant, we have to procced by first converting their masses into number of moles. Take each reactnat in turn and ask how many moles of product (urea) would be obtained if each were completely consumed. The reactant that gives the smaller number of moles of producet is the limiting reactant.
(ii) Convert the moles of product obtained to grams of product.
(iii) From the moles of product, calculate to grams fo excess reactant needed int he reaction. Then subtract this qunitity from the grams of the reactant available to find the quanity of the excess reactant remaining.

Answer 1

Step 1: Calculate the moles of each reactant.
Number of moels `(n) = ("Mass (g)")/("Molar mass" (g mol^(-1)))`
`? "mol" NH_(3) = 637.2g NH_(3) xx (1 "mol" NH_(3))/(17.00g NH_(3))`
`= 37.48 "mol" NH_(3)`
`? "mol" CO_(2) = 11.42g CO_(2) = 1142g CO_(2) xx (1 "mol" CO_(2))/(44.00 g CO_(2))`
`= 25.95 "mol" CO_(2)`
Step 2 : Convert the moles of each reactant to moles fo product `(NH_(2))_(2) CO`.
`37.48 "mol" NH_(3) xx (1 "mol" (NH_(2))_(2) CO)/(2 "mol" NH_(3)) = 18.74 "mol" (NH_(2))_(2) CO`
`25.95 "mol" CO_(2) xx (1 "mol" (NH_(2))_(2) CO)/(1 "mol" CO_(2)) = 25.95 "mol" (NH_(2))_(2) CO`
Thus, `NH_(3)`, yielding the smaller number of moles fo product must be the limiting reacatant producing `18.74 "mol" (NH_(2))_(2) CO` and `CO_(2)` is the exccss reactant , thus, some `CO_(2)` must be left unconsumed.
Step3 : Convert the moles of urea to grams fo urea.
`18.74 "mol" (NH_(2))_(2) CO xx (60.00g (NH_(2)) CO)/(1 "mol" (NH_(2))_(2) CO)`
`= 1124.4g (NH_(2))_(2) CO`
Step 4 : Convert the moles of area to grams of `CO_(2)` (the mass of `CO_(2)` needed to produce this amount of urea)
`18.74 "mol" (NH_(2))_(2) CO xx (1 "mol" CO_(2))/(1 mol (NH_(2))_(2) CO) xx (44.00 g CO_(2))/(1 "mol" CO_(2))`
`= 824.6 g CO_(2)`
Step 5 : Calculate the mass of `CO_(2)` left over.
As we started with `11.42 g CO_(2)` the mass of `CO_(2)` remaining un-consumed will be
`(1142 g CO_(2)) - (824.63g CO_(2)) = 317.4 g CO_(2)`