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

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