Urea `[(NH_2)_2CO]` is prepared by reacting ammonia with carbon dioxide. `2NH_3(g) + CO_2(g) (NH_2)_2CO (aq) +H_2O(l)` In one process, 637.2 g of NH3 are treated with 1142 g of `CO_2` .
How much excess reagent (in grams) is left at the end of the reaction ?

Explanation

Starting with 18.71 moles of `(NH_2)_2CO` , we can determine the mass of `CO_2` that reacted using the mole ratio from the balanced equation and the molar mass of `CO_2`
The conversion steps are
moles of `(NH_2)_2CO to` moles of `CO_2 to ` grams of `CO_2`
So that,
mass of `CO_2 ` reacted
` = 18.71 "mol" (NH_2)_2CO xx (1"mol" CO_2)/(1"mol"(NH_2)_2CO) xx (44.01 gCO_2)/(1"mol"CO_2)`
` = 823.4g`
The amount of `CO_2` remaining (in excess) is the difference between the initial amount (1142 g) and the amount reacted (823.4 g): mass of `CO_2` remaining = 1142 g - 823.4 g = 318.6 g ≈ 319 g