Question

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.

Answer 1

The balanced equation for the reaction between ammonia and oxygen is 4NH3 + 5O2 -> 4NO + 6H2O. To determine the amount of NO and H2O formed, we use the balanced equation to calculate the moles of products. With 2.19 mol NH3 and 4.93 mol O2, we can calculate the moles of NO and H2O produced.

Step-by-step explanation:

The balanced equation for the reaction between ammonia (NH3) and oxygen (O2) to form nitrogen monoxide (NO) and water (H2O) is:

4NH3 + 5O2 → 4NO + 6H2O

To determine the amount of NO and H2O that would form when 2.19 mol NH3 and 4.93 mol O2 react, we use the balanced equation to calculate the moles of products:

From the equation, we can see that 4 mol of NH3 produce 4 mol of NO. Therefore, 2.19 mol NH3 will produce (2.19 mol / 4 mol) * 4 mol of NO.

Similarly, 5 mol of O2 produce 4 mol of NO, so 4.93 mol O2 will produce (4.93 mol / 5 mol) * 4 mol of NO.

For the water, 4 mol of NH3 produce 6 mol of H2O, so (2.19 mol / 4 mol) * 6 mol of H2O will be produced. And for O2, (4.93 mol / 5 mol) * 6 mol of H2O will be produced.

Answer 2

NO would form 65.7 g.

H₂O would form 59.13 g.

Step-by-step explanation:

Given data:

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