Respuesta :
Answer: The amount of nitric acid produced is 30.81 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 nitrogen dioxide = 45.0 g
Molar mass of nitrogen dioxide = 46 g/mol
Putting values in equation 1, we get:
[tex]\text{Moles of nitrogen dioxide}=\frac{45g}{46g/mol}=0.978mol[/tex]
For the given chemical equation:
[tex]3NO_2(g)+H_2O(l)\rightarrow 2HNO_3(aq.)+NO(g)[/tex]
By Stoichiometry of the reaction:
3 moles of nitrogen dioxide produces 2 moles of nitric acid
So, 0.978 moles of nitrogen dioxide will produce = [tex]\frac{2}{3}\times 0.978=0.652mol[/tex] of nitric acid
- Now, calculating the mass of nitric acid from equation 1, we get:
Molar mass of nitric acid = 63 g/mol
Moles of nitric acid = 0.652 moles
Putting values in equation 1, we get:
[tex]0.652mol=\frac{\text{Mass of nitric acid}}{63g/mol}\\\\\text{Mass of nitric acid}=(0.652mol\times 63g/mol)=41.08g[/tex]
- To calculate the experimental yield of nitric acid, we use the equation:
[tex]\%\text{ yield}=\frac{\text{Experimental yield}}{\text{Theoretical yield}}\times 100[/tex]
Percentage yield of nitric acid = 75 %
Theoretical yield of nitric acid = 41.08 g
Putting values in above equation, we get:
[tex]75\%=\frac{\text{Experimental yield of nitric acid}}{41.08g}\times 100\\\\\text{Experimental yield of nitric acid}=\frac{75\times 41.08}{100}=30.81g[/tex]
Hence, the amount of nitric acid produced is 30.81 grams
