Question

The Ostwald process is used commercially to produce nitric acid, which is, in turn, used in many modern chemical processes. In the first step of the Ostwald process, ammonia is reacted with oxygen gas to produce nitric oxide and water. What is the maximum mass of H2O that can be produced by combining 87.4 g of each reactant?

4NH3 + 5O2 --> 4NO + 6 H2O

Answer 1

Given chemical reaction,

4NH3 + 5O2 -------> 4NO + 6H2O

No. of moles 4 5 4 6

Molar mass(g/mol) 17 32 30 18

Actual mass( in reaction) g 68 160 120 108

Mass used (g) 87.4 87.4 ? ?

From the table its clear that,

(4 mole NH3 + 5 mol O2) gives 6 moles of water(H2O)

Or (68 g NH3 + 160 g O2 = 228 g reactants) gives 108 g of water.

Then, (87.4 g NH3 +87.4 g O2 = 174.8 g reactants) gives say M g of water(H2O)

so we can write,

228 g of reactants = 108 g of H2O

174.8 g of reactants = 'M' g H2O

So, M = (174.8 x 108) / (228)

M = 82.8 g H2O.

The maximum mass of water (H2O) formed with 87.4 g of each reactant will be 82.8 g.

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We can also calculate mass of NO formed

Relation we can write as,

228 g of reactants = 120 g of NO

so 174.8 g reactants = 'Z' g of NO

Z = (120 x 174.8) / 228

Z = 92 g NO.