Nitric acid is made by a sequence of reactions, shown below. 4NH3(g) +5O2(g) = 4NO(g) + 6H2O(g)

2NO(g) + O2(g) = 2NO2(g)

3NO2(g) + H2O(g) = 2HNO3(g)+ NO(g)

If the first reaction occurs with 96.2% yield, the second reaction occurs with a 91.3% yield and the third reaction proceeds with a 91.4% yield, calculate the following:

(a) The grams of nitric acid produced from 1216 grams of ammonia. (b) The percent yield for the overall process

(a) In this case, by starting with 1216 grams of ammonia, we can firstly compute the yielded moles of NO in the first reaction considering the given yield as a fraction (0.962):

[tex]n_{NO}=1216 g NH_3*\frac{1molNH_3}{17gNH_3}*\frac{4molNO}{4molNH_3}*0.962=68.81molNO[/tex]

Next, in the second chemical reaction we compute the yielded moles of NO₂ with the 91.3-percent:

[tex]n_{NO_2}=68.81molNO*\frac{2molNO_2}{2molNO}*0.913=62.82molNO_2[/tex]

Finally, for the percent yield of the last chemical reaction and the molar mass of nitric acid (63 g/mol) we compute the yielded grams of nitric acid:

[tex]m_{HNO_3}=62.82molNO_2*\frac{2molHNO_3}{3molNO_2} *\frac{63gHNO_3}{1molHNO_3}*0.914\\ \\m_{HNO_3}=2412gHNO_3[/tex]

(b) In this case, we compute the moles of NO, NO₂ and the grams of nitric acid as well as the previous literal yet removing the percent yields since we are going to compute theoretical yields:

[tex]n_{NO}=1216 g NH_3*\frac{1molNH_3}{17gNH_3}*\frac{4molNO}{4molNH_3}=71.53molNO[/tex]

[tex]n_{NO_2}=71.53molNO*\frac{2molNO_2}{2molNO}=71.53molNO_2[/tex]

[tex]m_{HNO_3}=71.53molNO_2*\frac{2molHNO_3}{3molNO_2} *\frac{63gHNO_3}{1molHNO_3}*0.914=3004gHNO_3[/tex]

Thus, the overall percent yield is:

[tex]Y=\frac{2412g}{3004g} *100\%\\\\Y=80.3\%[/tex]

Best regards.