Question

1.28 g H2 is allowed to react with 10.4 g N2, producing 1.81 g NH3

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

Answer 1

N2 + 3H2 = 2NH3

Moles of N2 = mass / molar mass

= 10.4 /28

= 0.371

Moles of H2 = 1.28 / 2 = 0.64

1 mole N2 reacts with 3 mole of H2

So, 0.371 mole will react = 0.371*3 = 1.113 mole

But we have only 0.64 mole H2

So H2 is limiting reagent.

Now 3 mole H2 forms 2 mole NH3

0.64 mole will form = 0.43 mole

Mass of NH3 = 0.43*17 = 7.31 grams

Percentage yeild = (observed mass / calculated mass )*100

=( 1.81 / 7.31 )*100

= 24.76%

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