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

Respuesta :

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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  

=====  

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

======  

3. Calculate the mass of Al₂O₃  

Mass of Al₂O₃  = 1.00 × 101.95

Mass of Al₂O₃  = 1.02 g

Answer : The mass of [tex]Al_2O_3[/tex] produced will be, 101.96 grams

Explanation : Given,

Mass of Al = 54 g

Molar mass of Al = 27 g/mole

Molar mass of [tex]Al_2O_3[/tex] = 101.96 g/mole

First we have to calculate the moles of Al.

[tex]\text{Moles of }Al=\frac{\text{Mass of }Al}{\text{Molar mass of }Al}=\frac{54g}{27g/mole}=2mole[/tex]

Now we have to calculate the moles of [tex]Al_2O_3[/tex]

The given balanced chemical reaction is,

[tex]4Al+3O_2\rightarrow 2Al_2O_3[/tex]

From the balanced reaction we conclude that,

As, 4 moles of Al react to give 2 moles of [tex]Al_2O_3[/tex]

So, 2 moles of Al react to give [tex]\frac{2}{4}\times 2=1mole[/tex] of [tex]Al_2O_3[/tex]

Now we have to calculate the mass of [tex]Al_2O_3[/tex]

[tex]\text{Mass of }Al_2O_3=\text{Moles of }Al_2O_3\times \text{Molar mass of }Al_2O_3=1mole\times 101.96g/mole=101.96g[/tex]

Therefore, the mass of [tex]Al_2O_3[/tex] produced will be, 101.96 grams

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