Ammonia (nh3) can be produced by the reaction of hydrogen gas with nitrogen gas: 3h2 + n2 = 2nh3 a chemist reacts 2.00 mol h2 with excess n2. the reaction yields 0.54 mol nh3. what is the percent yield of the reaction? a) 25% b) 40% c) 60% d) 80%

The correct option is b.

Explanation: To calculate the percentage yield, we use the formula:

  ....(1)

For a given reaction:

0.54 moles of ammonia is formed.

So, amount of ammonia formed will be calculated using the formula:

     ....(2)

Molar mass of = 17.031g/mol

Amount of = 9.197g

Experimental yield : 9.197 g

By Stoichiometry of the reaction,

Here, limiting reagent is hydrogen gas because it limits the formation of product.

3 moles of hydrogen gas is producing 2 moles of ammonia

So, 2 moles of hydrogen gas will produce = moles of ammonia.

Amount of ammonia is calculated by using equation 2, we get:

Theoretical yield of ammonia  =22.65 grams

Now, putting values in equation 1, we get:

% yield=40.60 %

Hence, the correct option is b.

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