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

Sorry i dont know it is very hard to read and understand what u were trying to ask

Answer 2

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%