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 90.1 g of each reactant? 4NH3(g)+5O2(g)⟶4NO(g)+6H2O(g)

Answer 1

60.9 g

Step-by-step explanation:

The formula for the calculation of moles is shown below:

`moles = (Mass\ taken)/(Molar\ mass)`

Given: For
`NH_3`

Given mass = 90.1 g

Molar mass of
`NH_3` = 17.031 g/mol

Moles of
`NH_3` = 90.1 g / 17.031 g/mol = 5.2904 moles

Given: For
`O_2`

Given mass = 90.1 g

Molar mass of
`O_2` = 31.9988 g/mol

Moles of
`O_2` = 90.1 g / 31.9988 g/mol = 2.8157 moles

According to the given reaction:

`4NH_3_((g))+5O_2_((g))\rightarrow 4NO_((g))+6H_2O_((g))`

4 moles of
`NH_3` react with 5 moles of
`O_2`

1 mole of of
`NH_3` react with 5/4 moles of
`O_2`

5.2904 moles of
`NH_3` react with
`\frac {5}{4}* 5.2904` mole of
`O_2`

Moles of
`O_2` = 6.613 moles

Available moles of
`O_2` = 2.8157 moles

Limiting reagent is the one which is present in small amount. Thus,
`O_2` is limiting reagent. (2.8157 < 6.613)

The formation of the product is governed by the limiting reagent. So,

5 moles of
`O_2` gives 6 moles of
`H_2O`

1 mole of
`O_2` gives 6/5 moles of
`H_2O`

2.8157 moles of
`O_2` gives
`\frac {6}{5}* 2.8157` moles of
`H_2O`

Moles of water = 3.37884 moles

Molar mass of
`H_2O` = 18.0153 g/mol

Mass of
`H_2O` = Moles × Molar mass = 3.37884 × 18.0153 g = 60.9 g