Given the reaction: 4Al(s)+3O2(g)==2Al2O3(s) What is the minimum number of grams of oxygen gas required to produce 1.00 mole of aluminum oxide

4 Al(s)+3 O₂(g) ------------> 2 Al₂O₃(s)

3 moles O₂ -------------> 2 moles Al₂O₃
? moles O₂ --------------> 1.00 mole Al₂O3

O₂ = 1.00 * 3 / 2

O₂ = 3 / 2

= 1.5 moles of O₂

Molar mass O₂ = 32.0 g

1.5 * 32.0 = 48 g of O₂

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aliasger2709 aliasger2709 · Answer 2 · 2022-06-06T14:19:41+02:00

Grams of the gas can be estimated by multiplying the moles and the molar mass. To produce 1 mole of aluminum oxide, 48 gms of oxygen is required.

What is mass?

The mass of the substance is the product of the moles and the molar mass of the given substance.

The balanced chemical reaction is shown as:

4 Al (s) +3 O₂ (g) → 2 Al₂O₃ (s)

From the above reaction, it can be concluded that:

3 moles of oxygen = 2 moles of aluminum oxide

So, x moles of oxygen = 1.00 moles of aluminum oxide

x is calculated as:

x = 3 ÷ 2 = 1.5 moles

Mass of oxygen:

mass = moles × molar mass

= 1.5 × 32

= 48 gms

Therefore, 48 gms of oxygen is required to produce 1 mole of aluminum oxide.

Learn more about mass here:

https://brainly.com/question/17890371

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