Question

A 24.0 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

Answer 1

13.2%

Step-by-step explanation:

guy above is basically right except at the end he forgot to account for the 2 moles of NH3 from the reaction equation.

so, 3.85g/58.38g(100%)= 6.594724221 (2mol NH3)= 13.189= 13.2%

Answer 2

6.62%

Step-by-step explanation:

We have the following reaction

`N_2+3H_2 \longrightarrow 2NH_3`

A real yield of 3.85 g of
`NH_3` 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
`14 g/mol`

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

`14 g N \longrightarrow 1 mol N\\24.0g N \longrightarrow x \\x=((24)(1))/((14)) \\x= 1,71 mol N`

According to stoichiometric coefficients 1 mol of N produces 2 moles of
`NH_3`

`N_2+3H_2 \longrightarrow 2NH_3`

1.71 mol N how many moles of
`NH_3` will produce

`1 mol N\longrightarrow2 mol NH_3\\1.71 mol N \longrightarrow x \\x((1.71)(2))/(1) \\x= 3.42 mol NH_3`

Molar mass of
`NH_3` 17 g / mol

This means that in 1 mol of
`NH_3` there are 17 g.

`1 mol NH_3 \longrightarrow 17 g NH_3\\3.42 mol NH_3 \longrightarrow x\\x = ((3.42)(17))/(1) \\x= 58.14 g NH_3`

58.14g
`NH_3` corresponds to the theoretical yield of the reaction

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

`\%= (actual yield)/(theoretical yield) .100\%\\\%=(3.85)/(58.14). 100\%\\\%= 6.62\%`

The percent yield for this reaction is 6.62%