The Ostwald process is used commercially to produce nitric acid, which is, in tu

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 53.5 g of each reactant?

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 H20 that can be produced by combining 53.5 g of each reactant? 4NH, g)+ 50, 4N0(g) +6H2O Number g H2O

Explanation / Answer

we need to identify the limiting reagent . The reagent that is completely neutralised.

NH3

mass =53.5g

molar mass = 17g/mol

moles of NH3 = 53.5g /17g/mol = 3.147mol

O2

mass of O2 = 53.5g

molar mass = 32g/mol

moles = 53.5g/32g/mol = 1.672mol

According to stoichiometry :

4 moles of NH3 react with 5 moles of O2

3.147 moles of NH3 react 5/4 * 3.147 moles of O2 = 3.934 mol

So moles of O2 required = 3.934mol

but moles of O2 present 1.672 mol

Now, 5 mol of O2 react to give 6 moles of H2O

1.672mol of O2 react to give 6/5 * 1.672 =2.0064 moles of H2O

molar mass of H2O = 18g/mol

mass of H2O = 18g/mol* 2.0064 mol = 36.1152g= 36.11g of H2O

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