1.24 g H2 is allowed to react with 9.99 g N2 , producing 1.96 g NH3 .

Question

What is the theoretical yield in grams for this reaction under the given conditions?

What is the percent yield for this reaction under the given conditions?

Answer 1

Great Questions! The basic idea behind this problem is understanding: Which reactant is going to run out first?

Steps for all problems like these:

1) Convert everything to moles using the gram formula mass!! (Comparing apples to apples)

1.24 g H₂ * (1 mol H₂/2.016 g H₂) = 0.615 mol H₂

9.99 g N₂ * (1 mol N₂/28.02 g N₂) = 0.357 mol N₂

2) Now that we have comparable amounts of each reactant, we need to know the ratio of reactants that go into the reaction. This is found using the balanced equation

N₂ + H₂ -> NH₃ (This is not balanced, see how there is one extra nitrogen atom on the reactant side?)

N₂ + H₂ -> 2NH3 (Okay there is 2 nitrogen on each side, but now there is 6 hydrogen on the right?)

N₂ + 3H₂ -> 2NH₃ (Nice, now nitrogen and hydrogen are balanced and we know the moles of N₂ and H₂ to make NH3)

3) This is the tricky part. We need to figure out how much NH₃ we can make given one reactant is limited and the other doesn't matter (i.e we have so much available that we will never run out!)

Limited H₂: 0.615 mol H₂ * (2 mol NH₃/3 mol H₂) = 0.410 mol NH₃ (I made sure that NH₃ is in the numerator so that our final answer is mol NH₃, and 3 H₂ is in the denominator since we want to cancel that unit with 0.615 mol H_2. Where does the 2 mol NH₃/3 mol H₂ come from? The balanced equation! )

Limited N₂: 0.357 mol N₂ * (2 mol NH₃/1 mol N₂) = 0.750 mol NH₃

4) So we have two values for moles of NH₃, how do we choose? Simple: the smallest value is the answer! Why? It comes down to which ingredient (Reactant) you run out of first.

5) Finally convert the smaller value of product to grams (using the gram formula mass and we have our answer!

0.410 mol NH₃ * (17.031 g / mol NH₃ ) = 6.98 g (Usually okay to use question stem significant figures to report in answer, in this case, 3 sig figs from 1.24 g H₂)

6) Percent yield is the amount you obtained over the amount you CALCULATED you should have obtained.

(obtained/calculated) * 100 = (1.96 g NH₃/6.98 g NH₃) *100 = 28.1%

Answer 2

Welp, first we shouldn't get faked out by the amount of product given in the question. That will be the "actual" yield and we will get to that later.

First, it always starts with a balanced reaction. Let's balance the reaction of H2 and N2 reacting to produce NH3...

3 H2 + 1 N2 --> 2 NH3

Then, let's calculate the number of moles of each reactant from the masses given:

1.24 g H2 x 1 mol H2/2.02 g H2 = 0.614 mol H2

9.99 h N2 x 1mol N2/28.02 g N2 = 0.357 mol N2

Now we need to figure which reactant is the limiting reactant.

In the balanced reaction, we can see that the reaction requires 3 moles of H2 for every mole of N2. But, when we divide the number of moles of H2 given by the number of moles of N2 given, we do not have a 3:1 ratio, but something more like a 1.72 moles of H2 to 1 mole of N2 ratio (0.614/0.357).

So, H2 is the limiting reactant, and it will run out and completely diminish to zero. Always figure the limiting reactant as the reactant which will run out and go to zero, and therefore, limiting the reaction once it runs out. Let's now start with the number of moles of the limiting reactant H2 and calculate how many moles and then grams of NH3 can be produced. Remember the mole ratio in the balanced reaction!

0.614 mol H2 x 2 mol NH3/3 mol H3 = 0.409 mol NH3

And then we'll calculate the mass of NH3 theoretical. This will be our theoretical yield. We will use the molar mass of NH3 of 17.04 g/mol.

0.409 mol NH3 x 17.04 g NH3/1 mol NH3 = 6.98 g NH3.

The 6.98 g NH3 is the theoretical yield.

If the actual yield was only 1.96 g NH3, then the % yield is 1.96g NH3/6.98 g NH3 x 100& = 28.1 % yield!

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Mike