Answer :
The question is incomplete, here is the complete question:
Urea (CH₄N₂O) is a common fertilizer that can be synthesized by the reaction of ammonia (NH₃) with carbon dioxide as follows: 2NH₃(aq) + CO₂(aq) → CH₄N₂O(aq) + H₂O(l) In an industrial synthesis of urea, a chemist combines 135.9 kg of ammonia with 211.4 kg of carbon dioxide and obtains 178.0 kg of urea.
Determine the limiting reactant. (express your answer as a chemical formula)
Answer: The limiting reactant is ammonia [tex](NH_3)[/tex]
Explanation:
To calculate the number of moles, we use the equation:
[tex]\text{Number of moles}=\frac{\text{Given mass}}{\text{Molar mass}}[/tex] .....(1)
- For ammonia:
Given mass of ammonia = 135.9 kg = 135900 g (Conversion factor: 1 kg = 1000 g)
Molar mass of ammonia = 17 g/mol
Putting values in equation 1, we get:
[tex]\text{Moles of ammonia}=\frac{135900g}{17g/mol}=7994.12mol[/tex]
- For carbon dioxide gas:
Given mass of carbon dioxide gas = 211.4 kg = 211400 g
Molar mass of carbon dioxide gas = 44 g/mol
Putting values in equation 1, we get:
[tex]\text{Moles of carbon dioxide gas}=\frac{211400g}{44g/mol}=4804.54mol[/tex]
The given chemical reaction follows:
[tex]2NH_3(aq.)+CO_2(aq,)\rightarrow CH_4N_2O(aq.)+H_2O(l)[/tex]
By Stoichiometry of the reaction:
2 moles of ammonia reacts with 1 mole of carbon dioxide
So, 7994.12 moles of ammonia will react with = [tex]\frac{1}{2}\times 7994.12=3997.06mol[/tex] of carbon dioxide
As, given amount of carbon dioxide is more than the required amount. So, it is considered as an excess reagent.
Thus, ammonia is considered as a limiting reagent because it limits the formation of product.
Hence, the limiting reactant is ammonia [tex](NH_3)[/tex]