A 45.9 g sample of a metal is heated to 95.2°c and then placed in a calorimeter containing 120.0 g of water (c = 4.18 j/g°c) at
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Answer:
The answer is
<h2>219.5 mL</h2>Explanation:
The volume of a substance when given the density and mass can be found by using the formula
From the question
mass = 4500 g
density = 20.5 g/cm³
We have
We have the final answer as
<h3>219.51 mL</h3>Hope this helps you
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