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

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

3H2(g) + N2(g) —> 2NH3(g)

1.44g H2 is allowed to react with 10.5 g N2, producing 1.33g NH3.

Also, what is the percent yield for this reaction under the given conditions?

Any help on this would be much appreciated.

To answer the question, you must first establish the limiting reagent. to do that, convert the reactants to moles.

1.44/2.014 = 0.715 mol H2

10.5/28.013 = 0.375 mol N2

since you need 3 times the H2 as you do N2, H2 is the limiting reactant (0.715 mol of H2 will react with 0.238 mol N2)

following the chemical equation, you can produce 0.476 mol of NH3, which is 0.476*17.028= 8.11 grams, which is the theoretical yield.

the percent yield is the actual yield divided by the theoretical yield, in this case, 1.33/8.11 = 16.4%

Source(s): Way too much chemistry. note that I didn’t observe the proper number of significant figures throughout, but the answer has the proper number and is correct.

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

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