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When 1.57 mol of [tex]O_2[/tex] reacts with [tex]H_2[/tex] to form [tex]H_2O[/tex], how many moles of [tex]H_2[/tex] are consumed in the process?

Answer :

The reaction showing the synthesis of water from its elements:

2H₂ + O₂ --> 2H₂O

One mole of oxygen combines with 2 moles of hydrogen to form 1 mole of H₂O

If 1 mole O₂ consumes 2 moles H₂

then 1.57 moles of O₂ consumes = 1.57 x 2 moles of H₂

= 3.14 moles of H₂

Therefore, 3.14 moles of H₂ are consumed in the process.


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Rewritten by : Barada

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.