4. One of the steps in the manufacture of nitric acid is the combustion of ammon

Question

4. One of the steps in the manufacture of nitric acid is the combustion of ammonia with oxygen to produce nitrogen dioxide: NH3(g) + O2(g) right arrow NO2(g) + H2O(g) If 20,000 kg of nitrogen dioxide are needed in one of the reaction steps, how much ammonia would be required if the reaction occurred in only 72% yield? 5 . Hydrogen cyanide, an extremely toxic gas, is produced according to the reaction CH4(g) + O2(g) + NH3(g) right arrow HCN(g) + H2O(g) If the reaction occurs in 87% yield, how much each of ammonia and methane in grams will be required to produce 54.0 g of HCN?

Explanation / Answer

7)

from equation 1 mol HF = 1 mol UO2

a. No of mol of UO2 = 0.5*10^3/270.03 = 1.852 mol

mass of HF reacted = 1.852*20 = 37.04 grams.

b. No of moles of UO2 = 850/270.03 = 3.15 mol

from equations

1 mol UO2 = 1 mol of UF4 = 1 mol of UF6

mass of UF6 produced = 3.15*352.02 = 1108.86 grams

8) NH3 + O2 ----> NO2 + H2O

nO of mol of NO2 required = 20000*10^3/46 = 434782.61 mol

from equation 1 mol NO2 = 1 MOL NH3

so that

no of mol of NH3 required = 434782.61*100/72 = 603864.736 mol

mass of NH3 must be Taken = 603864.736*17 = 10265.7 kg

Related Questions

Navigate

• Browse (All)
• Browse 4
• Subjects

---

---

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.