The first step in the Ostwald process (used to produce nitric acid) is the forma

Question

The first step in the Ostwald process (used to produce nitric acid) is the formation of nitric oxide from ammonia and oxygen.

The first step in the Ostwald process (used to produce nitric acid) is the formation of nitric oxide from ammonia and oxygen. 4NH,(g) +502(g) 4N0(g) + 6H2O(g) What is the stoichiometric reactant ratio? (mol NH3 needed to react with 1 mol of O2) Number mol NH mol O2 0.8 How many moles of NO are formed from the reaction of 1 mol NH3? Number mol NO If the feed going to the reactor contains 0.640 mol NH3/mol O2, what is the limiting reactant, and the percent excess of the other reactant? Number O NH3 Limiting Reactant: % excess (Scroll down for more questions.)

Explanation / Answer

Answer – In this one there are is give reaction

4NH3(g) +5 O2(g) -----> 4 NO(g) + 6H2O(g)

The stoichiometric ratio of reactant, NH3 to O2 = 4/5 = 0.8

From the reaction

4 moles of NH3(g) = 4 moles of NO(g)

So, 1 moles of NH3(g) = ?

= 1 moles of NO(g)

Now we are given the 0.640 moles of NH3 for 1 moles of O2

We know

Mole ratio between NH3 and NO

So 0.640 moles of NH3 = 0.640 moles of NO

And from O2

5 moles of O2 = 4 moles of NO

So, 1 moles of O2 = ?

= 0.80 moles of NO

So, NH3 is limiting reactant

Excess moles of O2

From the reaction mole ratio between NH3 and O2 is 0.8

So, 4 moles of NH3 = 5 moles of O2

So for 0.640 moles of NH3 = ?

= 0.80 moles of O2

So, excess moles of O2 = 1.00-0.80 = 0.2 moles

So, percent of excess = 0.20 /1.0 *100 %

                                   = 20 %

Now , mass of NO = 1350 lbm = 612349.7 g

Moles of NO = 612349.7 g / 30.006 g/mol

                       = 20370.9 moles

So, mole ratio is 1:1

So, mole of NH3 = 20370.9 moles

Mass of NH3 = 20370.9 *17.0307 g/mol

                      = 346930.9 g

                      = 764.85 lbm

So, mass flow rate = 764.85 /24 hr = 31.89 lb m NH3 / hr

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