Respuesta :
Answer: Mass of [tex]NH_3[/tex] used is 1730812.5 grams.
Explanation: Ostwald's process is the process of the production of nitric acid from ammonia. Steps for the process are:
[tex]4NH_3(g)+5O_2(g)\rightarrow 4NO(g)+6H_2O(l)[/tex]
[tex]2NO(g)+O_2(g)\rightarrow 2NO_2(g)[/tex]
[tex]2NO_2(g)+H_2O(l)\rightarrow HNO_3(aq.)+HNO_2(aq.)[/tex]
Total mass of [tex]HNO_3[/tex] produced = 1.81 tons
Using conversion factor, we get
Total mass of [tex]HNO_3[/tex] produced = (1.81 × 2000 × 453.6)g = 1642032 g
Moles of [tex]HNO_3[/tex] is calculated by using the formula:
[tex]\text{Number of Moles}=\frac{\text{Given mass}}{\text{Molar mass}}[/tex] ....(1)
Molar mass of nitric acid = 63 g/mol
[tex]\text{Number of Moles}=\frac{1642032g}{63g/mol}=26064moles[/tex]
We are given that the percentage yield is 80 % in every step, then
Moles of [tex]NO_2[/tex], theorectically (From Step-III) = 2 × 26064 = 52128 moles
Actual moles of [tex]NO_2=52128\times \frac{100}{80}=65160moles[/tex]
Now, theoretical moles of NO (in Step-II) = Actual moles of [tex]NO_2[/tex]
Theoretical moles of NO = 65160 moles
Actual moles of NO = [tex]65160\times \frac{100}{80}=81450moles[/tex]
Now, theoretical moles of [tex]NH_3[/tex] (in Step-I) = Actual moles of NO
Theoretical moles of [tex]NH_3[/tex] = 81450 moles
Actual moles of [tex]NH_3[/tex] = [tex]81450\times \frac{100}{80}=101812.5moles[/tex]
To calculate the mass of [tex]NH_3[/tex], we use equation 1, we get
Molar mass of [tex]NH_3[/tex] = 17 g/mol
[tex]\text{Mass of }NH_3\text{ used}=101812.5mol\times 17g/mol=1730812.5g[/tex]
