Answer :
Answer: The mass of [tex]Al_2O_3[/tex] is 4.7 grams
Explanation:
To calculate the number of moles, we use the equation:
[tex]\text{Number of moles}=\frac{\text{Given mass}}{\text{Molar mass}}[/tex] .....(1)
For aluminium:
Given mass of aluminium= 2.5 g
Molar mass of aluminium = 27 g/mol
Putting values in equation 1, we get:
[tex]\text{Moles of aluminium}=\frac{2.5g}{27g/mol}=0.092mol[/tex]
For oxygen gas:
Given mass of oxygen gass = 2.5 g
Molar mass of oxygen gas = 32 g/mol
Putting values in equation 1, we get:
[tex]\text{Moles of oxygen gas:}=\frac{2.5g}{32g/mol}=0.078mol[/tex]
The chemical equation for the reaction of aluminium and oxygen gas follows:
[tex]4Al+3O_2\rightarrow 2Al_2O_3[/tex]
By Stoichiometry of the reaction:
4 moles of aluminium react with = 3 moles of oxygen
So, 0.092 moles of aluminium react with = [tex]\frac{3}{4}\times 0.092=0.069mol[/tex] of oxygen
As, given amount of oxygen is more than the required amount. So, it is considered as an excess reagent.
Thus, aluminium is considered as a limiting reagent because it limits the formation of product.
By Stoichiometry of the reaction:
4 moles of aluminium produces 2 moles of aluminium oxide
So, 0.092 moles aluminium produce = [tex]\frac{2}{4}\times 0.092=0.046moles[/tex] of aluminium oxide
Now, calculating the mass of aluminium oxide[tex]{\text{Mass of of aluminium oxide}}=moles\times {\text{Molar mass of aluminium oxide}}=0.046mol\times 102g/mol=4.7g[/tex]
Hence, the mass of [tex]Al_2O_3[/tex] is 4.7 grams