1) 4NH3(g)+5O2(g)>4NO(g)+6H2O(g)
2) 2NO(g)+O2(g)>2NO2(g)
3)3NO2(g)+H2O(l)>2HNO3(aq)+NO(g)
how much nitric oxide can be produced from a mixture of 5.7E6g of ammonia and 5.7E7g of O2
urgent and please explain
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The Ostwald Process for the Commercial Production of Nitric Acid Involves the Following Three Steps?
January 1st, 2011 
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Posted in Muscle Building Supplement 
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2 no2(g) 3 no2(g) h2o(l) ? 2 hno3(l) no(g) calculate delta h, d, the ostwald process for the commercial production of nitric acid involves three steps. answer, the ostwald process for the commercial production of nitric acid involves three steps.4 nh3(g) 5 o2(g) 4 no(g) 6 h2o(g)2 no(g) o2(g) 2 no2(g)3 no2(g) h2o(l) 2 hno3(l) no(g)(a) calculate ?h°, ?s°, ?g°,, the production of nitric acid by the ostwald process, third step of the ostwald process, Three, what is the ostwaldt process, yield in each step is 90% how much nitric acid will produce from nh3
First you need to determine your limiting reagent in the first step.
Molecular mass of NH3 = 17 g/mol, O2 = 32 g/mol
Mol of NH3 = 5,700,000 g/17 g/mol
Mol of NH3 = 3.35E5 mol
Mol of O2 = 57,000,000 g/32 g/mol
Mol of O2 = 1.78E6 mol
So based on the first equation (1), NH3 is the limiting reagent.
For every 4 mol of NH3, one gets 4 mol of NO. So the first reaction produced 3.35E5 mol of NO.
The second reaction produces 2 mol of NO2 for every 2 mol of NO used, so the second reaction produced 3.35E5 mol of NO2.
The third reaction produces 2 mol of HNO3 for every 3 mol of NO2 used, so this final reaction will produce 2/3 x 3.35E5 mol of HNO3 = 2.23E5 mol.
Then use molecular mass of HNO3 to calculate how much of it (mass) that you made.
HNO3 molecular mass = 63 g/mol
mass = mol x molar mass
mass = 2.23E5 x 63 g/mol
mass = 1.41E7 g of HNO3
Of course this assumes 100% yields in all the above noted reactions.