Question

6) (12 points) Ammonia and oxygen react to produce nitrogen gas and water vapor according to the reaction below. In a particular experiment, 3.92g of $N_2$ are produced by the reaction of 7g of $NH_3$ with 7g of $O_2$. What is the % yield of $N_2$ in this experiment? $NH_3(g) + O_2(g) \rightarrow N_2(g) + H_2O(g)$ (NOT BALANCED)

          6) (12 points) Ammonia and oxygen react to produce nitrogen gas and water vapor according to the reaction below.
In a particular experiment, 3.92g of $N_2$ are produced by the reaction of 7g of $NH_3$ with 7g of $O_2$. What is the % yield of $N_2$ in this experiment?
$NH_3(g) + O_2(g) \rightarrow N_2(g) + H_2O(g)$ (NOT BALANCED)

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6) (12 points) Ammonia and oxygen react to produce nitrogen gas and water vapor according to the reaction below.
In a particular experiment, 3.92g of N2 are produced by the reaction of 7g of NH3 with 7g of O2. What is the % yield of $N2$ in this experiment?
NH3(g) + O2(g) → N2(g) + H2O(g) (NOT BALANCED)

Added by Jose Antonio W.

Chemistry: Structure and Properties

Nivaldo Tro 2nd Edition

Instant Answer

Solved by Expert Chareen Guzman

05/03/2023

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Step 1: Calculate the moles of N2 and H2 used in the reaction. Show more…

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Yield of N in this experiment: NH3(g) N2(g) + 3H2(g) (BALANCED)

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Transcript

00:01 Hi, so we are to calculate the theoretical yield and then the percent yield of the reaction.

00:06 So we have here the balance equation is already given.

00:09 So let's identify first the limiting reactant in our reaction.

00:15 We'll do that by converting the given mass of the reactants to moles.

00:19 Number of moles of nh3, if we have 74 .2, this is kilograms, convert this to grams because we will be using the molar mass in terms of grams.

00:31 One kilogram is equivalent to 1000 grams.

00:34 The molar mass of nh3 is 74 .04 grams here and then moles in the numerator.

00:42 Now we cancel kilograms grams and we'll get 4354 .46.

00:50 This is moles of nh3 and then for co2, if we have 105 kilograms, convert to grams.

01:00 One kilogram is 1000 grams and then the molar mass of co2 is 44 .01 grams here and then moles in here.

01:11 Now we cancel kilograms, grams, grams and we'll get 2385 .82.

01:20 This is moles of co2.

01:23 Now we will solve for the mole ratio.

01:27 So mole ratio in this case is the number of moles of the reactant divided by the coefficient of that reactant from the balance equation.

01:35 So let's start with ammonia.

01:37 We have 4354 .46 divided by the coefficient of ammonia from the balance equation which is 2.

01:44 So we'll have 2177 .23.

01:50 For co2, we have 2385 .82 divided by 1.

01:55 So we'll have 2385 .82 also.

01:59 So whichever gives the smallest mole ratio will be the limiting reactant.

02:04 So that means ammonia is our limiting reactant here, our reactant.

02:10 So why did we have to identify the limiting reactant first? because the amount of the limiting reactant will be the basis on the amount of the products formed.

02:17 Now we could calculate for the theoretical yield which is the mass of urea.

02:24 That's ch4n2o...

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