A24.0 g sample of nitrogen gas reacts with an excess of hydrogen gas to give an actual yield of 3.85 g nh3. what is the percent yield for this reaction given the reaction: n2(
g. 3h2(
g. --> 2nh3(g

6.62%

Explanation:

We have the following reaction

A real yield of 3.85 g of is obtained

To calculate the reaction yield in percentage we must calculate the mass that theoretically we must obtain

The reaction occurs in excess of hydrogen and 24.0g of N this means that when the nitrogen is finished the reaction will end

Nitrogen is our limit reagent

Molar mass of nitrogen

This means that in 1 mol of nitrogen there is 14 g.

To calculate the moles of nitrogen in 24 g N we apply a simple rule of three

According to stoichiometric coefficients 1 mol of N produces 2 moles of

1.71 mol N how many moles of will produce

Molar mass of 17 g / mol

This means that in 1 mol of there are 17 g.

58.14g corresponds to the theoretical yield of the reaction

To calculate the percentage yield of the reaction we use the following formula

The percent yield for this reaction is 6.62%