Respuesta :
Answer: The percent yield of the reaction is 91.20 %.
Explanation:
To calculate the number of moles, we use the equation:
[tex]\text{Number of moles}=\frac{\text{Given mass}}{\text{Molar mass}}[/tex]
....(1)
Given mass of nitrogen gas = 140 g
Molar mass of nitrogen gas = 28 g/mol
Putting values in above equation, we get:
[tex]\text{Moles of nitrogen gas}=\frac{140g}{28g/mol}=5mol[/tex]
For the given chemical reaction:
[tex]N_2(g)+3H_2(g)\rightarrow 2NH_3(g)[/tex]
As, hydrogen is present in excess. Thus, it is considered as an excess reagent. Nitrogen gas is considered as limiting reagent because it limits the formation of products.
By Stoichiometry of the reaction:
1 mole of nitrogen gas produces 2 moles of ammonia.
So, 5 moles of nitrogen gas will produce = [tex]\frac{2}{1}\times 5=10mol[/tex] of ammonia.
Now, calculating the mass of ammonia from equation 1, we get:
Molar mass of ammonia = 17 g/mol
Moles of ammonia = 10 moles
Putting values in above equation, we get:
[tex]10mol=\frac{\text{Mass of ammonia}}{17g/mol}\\\\\text{Mass of ammonia}=170g[/tex]
To calculate the percentage yield of the reaction, we use the equation:
[tex]\%\text{ yield}=\frac{\text{Experimental yield}}{\text{Theoretical yield}}\times 100[/tex]
Experimental yield of ammonia = 155 g
Theoretical yield of ammonia = 170 g
Putting values in above equation, we get:
[tex]\%\text{ yield of ammonia}=\frac{155g}{170g}\times 100\\\\\% \text{yield of ammonia}=91.20\%[/tex]
Hence, the percent yield of the reaction is 91.20 %.
