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Answer :
Final answer:
By using stoichiometry and the molar masses of aluminium and aluminium oxide, we can calculate that 73.34 grams of aluminium oxide can be produced from 38.8 grams of aluminium reacting with oxygen.
Explanation:
To determine how many grams of aluminium oxide, Al2O3, can be formed from the reaction of 38.8 g of aluminium with oxygen, we use the given balanced chemical equation:
4 Al (s) + 3 O2(g) → 2 Al2O3(s)
First, we convert the mass of aluminium to moles using its molar mass (26.98 g/mol):
38.8 g Al × (1 mol Al / 26.98 g Al) = 1.439 moles Al
The stoichiometry of the reaction tells us that 4 moles of Al produce 2 moles of Al2O3, so we set up a ratio:
(1.439 moles Al) × (2 moles Al2O3 / 4 moles Al) = 0.7195 moles Al2O3
Finally, we convert moles of Al2O3 to grams using its molar mass (101.96 g/mol):
(0.7195 moles Al2O3) × (101.96 g Al2O3/mol) = 73.34 g Al2O3
Therefore, 73.34 grams of aluminium oxide can be formed by the reaction of 38.8 g of aluminium with oxygen.
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Answer: 50. 4g
Explanation:
First calculate number of moles of aluminium in 38.8g
Moles = 38.8g/ 26.982mol/g
= 1.44mol
By looking at the balance equation you can see that 4 moles of aluminium produce 2 moles of aluminium oxide.
4 = 2
1.4 = x
Find the value of x
x= (1.4×2)/4= 0.72 mol
0.72 moles of aluminium oxide are produced from 38.8g of aluminium
Now find the mass of aluminium produced.
Mass = moles × molar mass
= 0.72mol × 69.93 mol/g
= 50.4g
