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

Answer:

(a) $m_{HNO_3}=2412gHNO_3$

(b) $Y=80.3\%$

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):

$n_{NO}=1216 g NH_3*\frac{1molNH_3}{17gNH_3}*\frac{4molNO}{4molNH_3}*0.962=68.81molNO$

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

$n_{NO_2}=68.81molNO*\frac{2molNO_2}{2molNO}*0.913=62.82molNO_2$

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:

$m_{HNO_3}=62.82molNO_2*\frac{2molHNO_3}{3molNO_2} *\frac{63gHNO_3}{1molHNO_3}*0.914\\ \\m_{HNO_3}=2412gHNO_3$

(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:

$n_{NO}=1216 g NH_3*\frac{1molNH_3}{17gNH_3}*\frac{4molNO}{4molNH_3}=71.53molNO$

$n_{NO_2}=71.53molNO*\frac{2molNO_2}{2molNO}=71.53molNO_2$

$m_{HNO_3}=71.53molNO_2*\frac{2molHNO_3}{3molNO_2} *\frac{63gHNO_3}{1molHNO_3}*0.914=3004gHNO_3$

Thus, the overall percent yield is:

$Y=\frac{2412g}{3004g} *100\%\\\\Y=80.3\%$

Best regards.