Verified answer

anytime you get a question like "how much of this or that is formed" or what's the % yield. think of these steps. And you probably want to write them down somewhere.

1) write a balanced equation.

2) convert everything given to MOLES

3) determine limiting reagent

4) convert moles limiting reagent to moles other species via the coefficients of the balanced equation

5) convert moles back to mass. This is "theoretical mass" aka "theoretical yield"

6) % yield = actual measured mass / theoretical mass x 100%

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your problem...

*** 1 ***

N2 + 3 H2 ---> 2 NH3 is balanced.

you should always double check any reactions you are given btw....

*** 2 ***

moles N2 = 100.0 g N2 x (1 mole / 28.01 g N2) = 3.570 moles N2

*** 3 ***

since you were given "XS H2", you get to assume N2 is limiting. sometimes though, you have to compare the ratios to the coefficients of the balanced equation. example..

since 1 mole N2 reacts with 3 moles H2...therefore...

3.570 moles N2 x (3 moles H2 / 1 mole N2) = 10.71 moles H2

if you have more than than, N2 is limiting, if you have less H2 than 10.71 moles, H2 is limiting.

*** 4 ***

from the balanced equation, 1 mole N2 ---> 2 moles NH3.. therefore

3.570 moles N2 x (2 moles NH3 / 1 mole N2) = 7.140 moles NH3

*** 5 ***

7.140 moles NH3 x (17.03 g NH3 / mole NH3) = 121.6 g NH3

*** 6 ***

% yield = 34.0 g / 121.6 g x 100% = 28.0%

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quesitons?

Convert 100.0 g of nitrogen to mols:

Nitrogen is 14.0067 g/mol

N2 thus is 28.0134 g/mol

100.0 g / 28.0134 = 3.570 mols

Convert mols of N2 to mols of NH3

There are 2 mols of NH3 for every mol of N2

So: 3.570 mols of N2 * 2 = 7.140 mols NH3

Convert mols of NH3 to grams of NH3

NH3 = 17.03052 g/mol

Multiply 7.140 * 17.03052 = 121.6 grams NH3

Divide the actual yeild (34.0) by the theoretical yield (121.6)

34.0 g / 121.6 g = 0.2796

Multiply by 100

= 27.96%