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 54.5 g of each reactant?

Question

Alexzander Haney

Chemistry

18 January, 08:43

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 54.5 g of each reactant?

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Milo Huffman · Answer 1

The equation describing the reaction between ammonia and oxygen is described as follows:

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

From the periodic table:

mass of oxygen = 16 grams

mass of hydrogen = 1 gram

mass of nitrogen = 14 grams

Therefore,

mass of NH3 = 14 * 3 (1) = 17 grams

mass of O2 = 2*16 = 32 grams

mass of H2O = 2 (1) + 16 = 18 grams

From the balanced equation above:

4 (17) = 68 grams of NH3 are required to react with 160 grams of oxygen to produce 108 grams of water

In this reaction, NH3 is the limiting reagent. This means that the reaction would stop once the amount of NH3 is consumed.

Therefore, we will base the calculations on the amount of NH3.

Now, we know that 68 grams of ammonia produce 108 grams of water. To know the amount of water produced from 54.5 grams of NH3, we will simply do a cross multiplication as follows:

amount of water = (54.5*108) / 68 = 86.5588 grams