QUESTION ❓✔️❓
• Urea [(NH2)2CO] is prepared by reacting ammonia with carbon dioxide.
2NH3(g) + CO2(g) ------>
(NH2)2CO (aq) + H2O(1)
( In one process, 637.2 g of NH3, are treated with 1142 g of CO2.)
a) Which of the two reactants is limiting reagent ?
2) Calculate the mass of (NH2)2CO formed?
3) How much excess reagent in (grams) is left at the end of the reaction?
Answers
Answer :-
(1) NH₃ is limiting reagent .
(2) 1124.4 grams of urea is formed .
(3) 317.24 grams of excess reagent is left .
Explanation :-
2NH₃ (g) + CO₂ (g) → (NH₂)₂CO (aq) + H₂O (l)
Moles of ammonia (NH₃)
= Given mass/Molar mass
= 637.2/17
= 37.48 moles
Moles of carbon dioxide (CO₂)
= 1142/44
= 25.95 moles
Now according to the balanced equation, 2 moles of ammonia react with 1 mole of carbon dioxide to form urea and water . So, 37.48 moles of ammonia will react with :-
= 37.48/2
= 18.74 moles of CO₂
______________________________
So, NH₃ is the limiting reagent and will control the amount of product . CO₂ is the excess reagent as (25.95 - 18.74) = 7.21 moles of CO₂ will be left after reaction .
∵ 2 moles of NH₃ → 1 mole of urea
∴ 37.48 moles of NH₃ → 18.74 moles of urea
Mass of (NH₂)₂CO formed :-
= Number of moles × Molar mass
= 17.48 × 60
= 1124.4 g
Mass of CO₂ left :-
= No of moles left × Molar mass
= 7.21 × 44
= 317.24 g
kindly see attachment for ur amswer
too lengthy(~_^)
