Question

For the equation 4Al + 3O2 → 2Al2O3, how many grams of aluminum oxide can be produced from 54 grams of aluminum?

Answer 1

Answer : The mass of
`Al_2O_3` produced will be, 101.96 grams

Explanation : Given,

Mass of Al = 54 g

Molar mass of Al = 27 g/mole

Molar mass of
`Al_2O_3` = 101.96 g/mole

First we have to calculate the moles of Al.

`\text{Moles of }Al=\frac{\text{Mass of }Al}{\text{Molar mass of }Al}=(54g)/(27g/mole)=2mole`

Now we have to calculate the moles of
`Al_2O_3`

The given balanced chemical reaction is,

`4Al+3O_2\rightarrow 2Al_2O_3`

From the balanced reaction we conclude that,

As, 4 moles of Al react to give 2 moles of
`Al_2O_3`

So, 2 moles of Al react to give
`(2)/(4)* 2=1mole` of
`Al_2O_3`

Now we have to calculate the mass of
`Al_2O_3`

`\text{Mass of }Al_2O_3=\text{Moles of }Al_2O_3* \text{Molar mass of }Al_2O_3=1mole* 101.96g/mole=101.96g`

Therefore, the mass of
`Al_2O_3` produced will be, 101.96 grams

Answer 2

Answers:

1.02 g

Explanation:

We know we will need a balanced equation with masses and molar masses, so let’s gather all the information in one place.

M_r: 26.98 101.95

4Al + 3O₂ ⟶ 2Al₂O₃

Mass/g: 54

1. Calculate the moles of Al

Moles Al = 54 × 1/26.98

Moles Al = 2.00 mol Al

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2. Calculate the moles of Al₂O₃

The molar ratio is 2 mol Al₂O₃:4 mol Al

Moles of Al₂O₃ = 2.00 × 2/4

Moles of Al₂O₃ = 1.00 mol Al₂O₃

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3. Calculate the mass of Al₂O₃

Mass of Al₂O₃ = 1.00 × 101.95

Mass of Al₂O₃ = 1.02 g