Answer:
6.62%
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
\(14 g N \longrightarrow 1 mol N\\24.0g N \longrightarrow x \\x=\frac{(24)(1)}{(14)} \\x= 1,71 mol N\)
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
\(1 mol N\longrightarrow2 mol NH_3\\1.71 mol N \longrightarrow x \\x\frac{(1.71)(2)}{1} \\x= 3.42 mol NH_3\)
Molar mass of \(NH_3\) 17 g / mol
This means that in 1 mol of \(NH_3\) there are 17 g.
\(1 mol NH_3 \longrightarrow 17 g NH_3\\3.42 mol NH_3 \longrightarrow x\\x = \frac{(3.42)(17)}{1} \\x= 58.14 g NH_3\)
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
\(\%= \frac{actual yield}{theoretical yield} .100\%\\\%=\frac{3.85}{58.14}. 100\%\\\%= 6.62\%\)
The percent yield for this reaction is 6.62%