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 H2OH2O that can be produced by combining 79.3 g79.3 g of each reactant?
The first step reaction of Ostwald process:
4NH3 + 5O2 = 6H2O + 4NOGiven masses of reactants: m(NH3) = 79.3g
m(O2) = 79.3g
Calculated molar masses: M(NH3) = 17 g/mol
M(O2) = 32 g/mol M(H2O) = 18 g/mol Amounts of reactants: n = m/M
n(NH3) = 79.3/17 = 4.665 mol n(O2) = 79.3/32 = 2.47 mol
It is obviously (accourding to the chemical reaction coefficients) that 4 moles of NH3 react with 5 moles of O2 completely. But we have 4.665 and 2.47 moles respectively, so NH3 is in abundance. So, all calculations will be made using the amount of oxygen.
5 moles of O2 give 6 moles of H2O
2.47 moles of O2 give x moles of H2O x = 2.47×6/5 = 2.97 mol The mass of water formed is m = nM = 2.97×18 = 53.527 g
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