Question

Determine the number of moles of oxygen gas (O2) that are needed to react with 0.725 moles of aluminum (Al).

4Al (s) + 3O2 (g) ⟶ 2Al2O3 (s)

Answer 1

Given :

Oxygen gas ($O_2$) that are needed to react with 0.725 moles of aluminum (Al).

$4Al(s)+3O_2(g)->2Al_2O_3(s)$

Solution :

4 moles of Aluminium react with 3 moles of oxygen,

Let, moles of oxygen react with 0.725 moles of Al is x.

$x=\dfrac{0.725\times 3}{4}\\\\x=0.54375\ moles$

Hence, this is the required solution.

Answer 2

The number of moles that should be required to react is 0.54375 moles.

Calculation of the number of moles:

Since There are 0.725 moles of aluminum (Al)

And, there is the chemical equation 4Al (s) + 3O2 (g) ⟶ 2Al2O3 (s)

The above equation represent 4 moles of Aluminium should be react with 3 moles of oxygen.

So, the number of moles should be

= 0.725 * 3/4

= 0.54375 moles

Therefore, The number of moles that should be required to react is 0.54375 moles.

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