Respuesta :
2H₂₍g₎ + O₂ ₍g₎→ 2H₂O
138 mol H₂ × (2 mol H₂O ÷ 2 mol H₂)= 138 mol H₂O
64 mol O₂ × (2 mol H₂O ÷ 1 mol O₂)= 128 mol H₂O
128 mol H₂O
138 mol H₂ × (2 mol H₂O ÷ 2 mol H₂)= 138 mol H₂O
64 mol O₂ × (2 mol H₂O ÷ 1 mol O₂)= 128 mol H₂O
128 mol H₂O
Answer : The number of moles of water formed can be, 128 moles
Solution : Given,
Moles of hydrogen gas = 138 moles
Moles of oxygen gas = 64 moles
First we have to calculate the limiting and excess reagent.
The balanced chemical reaction will be,
[tex]2H_2+O_2\rightarrow 2H_2O[/tex]
From the balanced reaction we conclude that,
As, 2 moles of hydrogen gas react with 1 mole of oxygen gas
So, 138 moles of hydrogen gas react with [tex]\frac{138}{2}=69[/tex] moles of oxygen gas
But the given moles of oxygen gas is 64 moles.
That means oxygen gas is a limiting reagent and hydrogen gas is an excess reagent.
Now we have to calculate the moles of water.
From the reaction we conclude that,
As, 2 moles of oxygen react to give 1 mole of water
So, 64 moles of oxygen react to give [tex]64\times 2=128[/tex] moles of water
Therefore, the number of moles of water formed can be, 128 moles