Question

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

Answer 1

(a)
`m_(HNO_3)=2412gHNO_3`

(b)
`Y=80.3\%`

Step-by-step explanation:

Hello.

(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):

`n_(NO)=1216 g NH_3*(1molNH_3)/(17gNH_3)*(4molNO)/(4molNH_3)*0.962=68.81molNO`

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

`n_(NO_2)=68.81molNO*(2molNO_2)/(2molNO)*0.913=62.82molNO_2`

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:

`m_(HNO_3)=62.82molNO_2*(2molHNO_3)/(3molNO_2) *(63gHNO_3)/(1molHNO_3)*0.914\\ \\m_(HNO_3)=2412gHNO_3`

(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:

`n_(NO)=1216 g NH_3*(1molNH_3)/(17gNH_3)*(4molNO)/(4molNH_3)=71.53molNO`

`n_(NO_2)=71.53molNO*(2molNO_2)/(2molNO)=71.53molNO_2`

`m_(HNO_3)=71.53molNO_2*(2molHNO_3)/(3molNO_2) *(63gHNO_3)/(1molHNO_3)*0.914=3004gHNO_3`

Thus, the overall percent yield is:

`Y=(2412g)/(3004g) *100\%\\\\Y=80.3\%`

Best regards.