Respuesta :
Aluminum oxide (used as an adsorbent or a catalyst for organic reactions) forms when aluminum reacts with oxygen. 4al(s) + 3o2(g) ® 2al2o3(s) [balanced] a mixture of 82.49 g of aluminum ( = 26.98 g/mol) and 117.65 g of oxygen ( = 32.00 g/mol) is allowed to react. what mass of aluminum oxide ( = 101.96 g/mol) can be formed?
Answer : The mass of aluminium oxide formed can be 155.8 grams.
Solution : Given,
Mass of Al = 82.49 g
Mass of [tex]O_2[/tex] = 117.65 g
Molar mass of Al = 26.98 g/mole
Molar mass of [tex]O_2[/tex] = 32.00 g/mole
Molar mass of [tex]Al_2O_3[/tex] = 101.96 g/mole
First we have to calculate the moles of Al and [tex]O_2[/tex].
[tex]\text{ Moles of }Al=\frac{\text{ Mass of }Al}{\text{ Molar mass of }Al}=\frac{82.49g}{26.98g/mole}=3.057moles[/tex]
[tex]\text{ Moles of }O_2=\frac{\text{ Mass of }O_2}{\text{ Molar mass of }O_2}=\frac{117.65g}{32.00g/mole}=3.677moles[/tex]
Now we have to calculate the limiting and excess reagent.
The balanced chemical reaction is,
[tex]4Al(s)+3O_2(g)\rightarrow 2Al_2O_3(s)[/tex]
From the balanced reaction we conclude that
As, 4 mole of [tex]Al[/tex] react with 3 mole of [tex]O_2[/tex]
So, 3.057 moles of [tex]Al[/tex] react with [tex]\frac{3.057}{4}\times 3=2.293[/tex] moles of [tex]O_2[/tex]
From this we conclude that, [tex]O_2[/tex] is an excess reagent because the given moles are greater than the required moles and [tex]Al[/tex] is a limiting reagent and it limits the formation of product.
Now we have to calculate the moles of [tex]Al_2O_3[/tex]
From the reaction, we conclude that
As, 4 mole of [tex]Al[/tex] react to give 2 mole of [tex]Al_2O_3[/tex]
So, 3.057 mole of [tex]Al[/tex] react to give [tex]3.057\times \frac{2}{4}=1.528[/tex] moles 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[/tex]
[tex]\text{ Mass of }Al_2O_3=(1.528moles)\times (101.96g/mole)=155.8g[/tex]
Therefore, the mass of aluminium oxide formed can be 155.8 grams.
