Answer:
mH₂O = 6.4116 g
Explanation:
Let's write the given reaction:
2H₂ + O₂ -------> 2H₂O
The problem states that 5.7 g of oxygen reacts with excess hydrogen, hence, the limiting reagent is the oxygen. With this mass of oxygen, we can determine the moles, and then, the moles of water with the mole ratio:
moles = mass / atomic weight AW of O₂ = 16 g/mol
Replacing we have:
moles O₂ = 5.70 / (16 * 2)
moles O₂ = 0.1781 moles
According to the balanced reaction, 1 mole of Oxygen produces 2 moles of water, so we have a mole ratio 1:2, therefore the moles of water would be twice the moles of oxygen:
moles H₂O = 0.1781 * 2 = 0.3562 moles
Finally the mass of water can be calculated solving for the mass from the expression of moles, and using molecular mass of water:
m = moles * MM MM H₂O = 18 g/mol
m = 0.3562 * 18
<h2>
mH₂O = 6.4116 g</h2>
Hope this helps
Answer:
You have to remember this memory aid, diagonals and all.
You have to keep a tight tally of the electrons you’ve used so far so you don’t go over the number of electrons in the element you’re working on.
You have to remember how many electrons fit into each subshell (s, p, d, f).
It takes a lot of time, especially when the element has more than 20 electrons.
Explanation:
What are you talking about and what’s the question
