he 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 93.3 g of each reactant?

148.2 g of H20

Explanation:

Equation of reaction: 4NH3 + 5O2 ---> 4NO + 6H20

From the equation above, 4 moles of ammonia reacts with 5 moles of oxygen gas to produce 6 moles water.

Molar mass of NH3 = 17 g/mol;

Molar mass of O2= 32 g/mol;

Molar mass of H2O = 18 g/mol

First, we determine the limiting reactant:

4*17 g of NH3 reacts with 5*32 g of O2

Mass Ratio = 68 : 160

Therefore, NH3 is the limiting reactant.

68 g of NH3 reacts to produce 6* 18 g of H20 = 108 g of H2O

93.3 g of NH3 will react to produce (93.3 * 108)/68 g of H20 = 148.2 g of H2O

Therefore, the maximum amount of H2O produced = 148.2 g

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