Urea (H2 NCONH2) is used extensively as a nitrogen source in...

Urea $\left(\mathrm{H}_{2} \mathrm{NCONH}_{2}\right)$ is used extensively as a nitrogen source in fertilizers. It is produced commercially from the reaction of ammonia and carbon dioxide: $$2 \mathrm{NH}_{3}(g)+\mathrm{CO}_{2}(g) \longrightarrow \mathrm{H}_{2} \mathrm{NCONH}_{2}(s)+\mathrm{H}_{2} \mathrm{O}(g)$$ Ammonia gas at $223^{\circ} \mathrm{C}$ and $90 .$ atm flows into a reactor at a rate of $500 .$ L/min. Carbon dioxide at $223^{\circ} \mathrm{C}$ and $45$ atm flows into the reactor at a rate of $600 .$ L/min. What mass of urea is produced per minute by this reaction assuming $100 \%$ yield?

Urea $\left(\mathrm{H}_{2} \mathrm{NCONH}_{2}\right)$ is used extensively as a nitrogen source in fertilizers. It is produced commercially from the reaction of ammonia and carbon dioxide:
$$2 \mathrm{NH}_{3}(g)+\mathrm{CO}_{2}(g) \longrightarrow \mathrm{H}_{2} \mathrm{NCONH}_{2}(s)+\mathrm{H}_{2} \mathrm{O}(g)$$
Ammonia gas at $223^{\circ} \mathrm{C}$ and $90 .$ atm flows into a reactor at a rate of $500 .$ L/min. Carbon dioxide at $223^{\circ} \mathrm{C}$ and $45$ atm flows into the reactor at a rate of $600 .$ L/min. What mass of urea is produced per minute by this reaction assuming $100 \%$ yield?

Key Concepts

Reaction Stoichiometry

Reaction stoichiometry involves using the coefficients in a balanced chemical equation to understand the relationship between reactants and products. This concept allows one to calculate how much product can be produced from given amounts of reactants, ensuring that the law of conservation of mass is satisfied in a chemical reaction.

Limiting Reactant

The limiting reactant is the substance in a chemical reaction that is completely consumed first, thereby determining the maximum amount of product that can be formed. Understanding which reactant limits the reaction is essential for accurate yield predictions and efficient resource utilization.

Ideal Gas Law

The ideal gas law relates the pressure, volume, temperature, and number of moles of a gas, and it is crucial for converting volumetric flow rates under specified conditions into moles. This conversion is fundamental when dealing with gas-phase reactions in order to connect measured flow rates to stoichiometric calculations.

Reaction Yield Calculation

Reaction yield calculation involves determining the mass or number of moles of the product formed, based on the limiting reactant and the stoichiometric relationships from the balanced equation. This concept is important for predicting the efficiency of industrial processes and for scaling up chemical reactions.

Transcript

00:02 Hi there.

00:03 In problem 74, we are looking at the production of urea.

00:08 So first thing i want to do is copy down our balanced equation.

00:14 We start with ammonia and add carbon dioxide.

00:23 Both of these are gases to produce the solid urea, which is h2, n, c -o -n -c -o -n -h -2.

00:41 And the other product is water, which would also be a gas.

00:47 All right.

00:48 So it looks like we have some information provided to us for both the ammonia and the carbon dioxide.

00:59 And we would like to know what mass of urea we can make per minute.

01:05 So question mark grams per minute.

01:10 Starting off, we know that the ammonia is flowing in a rate.

01:20 Of 500 liters per minute.

01:30 And the carbon dioxide is flowing in at a rate of 600 liters per minute.

01:43 Right.

01:44 So as we look at this, this looks an awful lot like a limiting reactant stoichiometry problem, which in fact it is.

01:52 But instead of liters per minute, we are going to need to get two moles per minute before we can decide which of these is our limiting reactant.

02:04 Because both of these are at different pressures, so we cannot just compare them straight away.

02:10 We are going to have to use pv equals nrt on both of these and calculate the number of moles of the ammonia and the carbon dioxide.

02:26 All right, so let's get started with that.

02:29 Let's start with the ammonia first.

02:36 So if i'm using pv equals nrt, n is going to be, equal to pv divided by rt.

02:46 So let's solve for the number of moles of ammonia.

02:49 We are given that the gas is at 90 atmospheres and it's flowing at a volume of 500 liters per minute.

03:08 The universal gas constant that i want to use is the one that has atmospheres in it so that atmospheres will cancel.

03:14 It's going to be 0 .0821 liters times atmospheres.

03:20 Divided by mole kelvin.

03:27 And the temperature is 496 kelvin.

03:33 It was given to us in celsius of 223, so adding 273 to that gives us 496 kelvin.

03:42 Temperature must be in kelvin, so that it will cancel the kelvin in our universal gas constant.

03:50 Okay, liters will cancel, atmospheres will cancel, and is going to be expressed in terms of moles, per minute.

04:01 So calculating this out, i get 1 ,106 moles per minute.

04:09 I'm not going to round two significant figures right now.

04:12 We'll wait till the very end of the problem to take care of that.

04:16 And this was the nh3...

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