0.03154 mol H₂O is formed from 0.0636 g of hydrogen if we consider that O₂ is a limiting reagent.
The balanced equation is
[tex]2H_{2} (g)[/tex] + [tex]O_{2} (g)[/tex] → [tex]2H_{2} O (l)[/tex]
Here, we are calculating the moles [tex]H_{2}O (l)[/tex] from the moles of H₂ if we consider O₂ is a limiting reagent:
Given:
The mass of H₂ = 0.0636 g
We know that,
The moles of H₂ = (0.0636 g of H₂) × [tex]\frac{1.0 mol H_{2} }{2.016 g H_{2} }[/tex]
= 0.03154 mol H₂
Moles of H₂O (l) from H₂ = ( 0.03154 mol H₂) × [tex]\frac{2.0 mol of H_{2} O (l)}{2.O mol H_{2} }[/tex]
= 0.03154 mol H₂O
Therefore, 0.03154 mol H₂O is formed from 0.0636 g of hydrogen if we consider that O₂ is limiting reagent.
Learn more about limiting reagent here:
brainly.com/question/26905271
#SPJ4
