Here we have to get the moles of hydrogen (H₂) consumed to form water (H₂O) from 1.57 moles of oxygen (O₂)
In this process 3.14 moles of H₂ will be consumed.
The balanced reaction between oxygen (O₂) and hydrogen (H₂); both of which are in gaseous state to form water, which is liquid in nature can be written as-
2H₂ (g) + O₂ (g) = 2H₂O (l).
Thus form the equation we can see that 1 mole of oxygen reacts with 2 moles of hydrogen to form 2 moles of water.
So, 1.57 moles of oxygen will consume (1.57×2) = 3.14 moles of hydrogen to form water.
First, let's write down the balanced chemical reaction between the given reactants:
NO₂ + NO → N₂O + O₂
The Lewis structure of the main product is shown in the attached picture. To determine the formal charge of each element, the formula is as follows:
Formal Charge = Valence electrons - Non-bonding valence electrons - (Bonding electrons/2)
For the leftmost N:
Formal charge = 5 - 2 - 6/2 = 0
For the middle N:
Formal charge = 5 - 0 - 8/2 = 1
For O:
Formal charge = 6 - 6 - 2/2 = -1
The volumeof this sphere must be less than 0.7228
Answer:
The answer to your question is 50 moles of O₂
Explanation:
Balanced Chemical reactions
1.- N₂(g) + 3H₂ (g) ⇒ 2NH₃ (g)
2.- 4NH₃ (g) + 5O₂(g) ⇒ 4NO (g) + 6H₂O (l)
moles of N₂(g) = 20 moles
moles of O₂(g) = ?
Process
1.- Calculate the moles of NH₃
1 mol of N₂ ------------- 2 moles of NH₃
20 moles of N₂ --------- x
x = (20 x 2) / 1
x = 40 moles of NH₃
2.- Calculate the moles of O₂
4 moles of NH₃ -------------- 5 O₂
40 moles of NH₃ ------------ x
x = (40 x 5) / 4
x = 200 / 4
x = 50 moles of O₂