Bakit tinawag na pacific ring of fire ang mga bansa sa timog silangan asya at pasipiko? A. dahil sa aktibong bulkan B.dahil sa l
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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)
<u>Answer:</u> The limiting reactant is ammonia
<u>Explanation:</u>
To calculate the number of moles, we use the equation:
.....(1)
- <u>For ammonia:</u>
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:
- <u>For carbon dioxide gas:</u>
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:
The given chemical reaction follows:
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 = 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
We know that to relate solutions of with the factors of molarity and volume, we can use the equation: 
**NOTE: The volume as indicated in this question is defined in L, not mL, so that conversion must be made. However it is 1000 mL = 1 L.
So now we can assign values to these variables. Let us say that the 18 M
is the left side of the equation. Then we have:
We can then solve for
:
and 
or 
We now know that the total amount of volume of the 4.35 M solution will be 210 mL. This is assuming that the entirety of the 50 mL of 18 M is used and the rest (160 mL) of water is then added.
**NOTE: The volume as indicated in this question is defined in L, not mL, so that conversion must be made. However it is 1000 mL = 1 L.
So now we can assign values to these variables. Let us say that the 18 M
We can then solve for
We now know that the total amount of volume of the 4.35 M solution will be 210 mL. This is assuming that the entirety of the 50 mL of 18 M is used and the rest (160 mL) of water is then added.