Ammonia and oxygen react to form nitrogen monoxide and water. Construct your own balanced equation to determine the amount of NO and H 2 O that would form when 2.19 mol N H 3 and 4.93 mol O 2 react.

NO would form 65.7 g.

H₂O would form 59.13 g.

Explanation:

Given dа ta:

Moles of NH₃ = 2.19

Moles of O₂ = 4.93

Mass of NO produced = ?

Mass of produced H₂O = ?

Solution:

First of all we will write the balance chemical equation,

4NH₃ + 5O₂ → 4NO + 6H₂O

Now we will compare the moles of NO and H₂O with ammonia from balanced chemical equation:

NH₃ : NO NH₃ : H₂O

4 : 4 4 : 6

2.19 : 2.19 2.19 : 6/4 * 2.19 = 3.285 mol

Now we will compare the moles of NO and H₂O with oxygen from balanced chemical equation:

O₂ : NO O₂ : H₂O

5 : 4 5 : 6

4.93 : 4/5*4.93 = 3.944 mol 4.93 : 6/5 * 4.93 = 5.916 mol

we can see that moles of water and nitrogen monoxide produced from the ammonia are less, so ammonia will be limiting reactant and will limit the product yield.

Mass of water = number of moles * molar mass

Mass of water = 3.285 mol * 18 g/mol

Mass of water = 59.13 g

Mass of nitrogen monoxide = number of moles * molar mass

Mass of nitrogen monoxide = 2.19 mol * 30 g/mol

Mass of nitrogen monoxide = 65.7 g