Respuesta :
Answer: The theoretical yield of nitrogen monoxide is 4.23 grams
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 oxygen gas = 5.62 g
Molar mass of oxygen gas = 32 g/mol
Putting values in equation 1, we get:
[tex]\text{Moles of oxygen gas}=\frac{5.62g}{32g/mol}=0.176mol[/tex]
The chemical equation for the reaction of oxygen gas and ammonia is:
[tex]4NH_3+5O_2\rightarrow 4NO+6H_2O[/tex]
By Stoichiometry of the reaction:
5 moles of oxygen gas produces 4 moles of nitrogen monoxide
So, 0.176 moles of oxygen gas will produce = [tex]\frac{4}{5}\times 0.176=0.141[/tex] moles of nitrogen monoxide
Now, calculating the mass of nitrogen monoxide from equation 1, we get:
Molar mass of nitrogen monoxide = 30 g/mol
Moles of nitrogen monoxide = 0.141 moles
Putting values in equation 1, we get:
[tex]0.141mol=\frac{\text{Mass of nitrogen monoxide}}{30g/mol}\\\\\text{Mass of nitrogen monoxide}=(0.141mol\times 30g/mol)=4.23g[/tex]
Hence, the theoretical yield of nitrogen monoxide is 4.23 grams
