Answer: -
164.5 grams of oxygen is present in 350 g of Al(C₂H₃O₂)₃
Explanation: -
Molar mass of Al(C₂H₃O₂)₃ = 27 x 1 + 12 x 6 + 1 x 9 + 16 x 6
= 204 g
Mass of oxygen present in 204 g of Al(C₂H₃O₂)₃ = 16 x 6 = 96 g
204 g of Al(C₂H₃O₂)₃ contains 96 g of oxygen.
350 g of Al(C₂H₃O₂)₃ contains [tex] \frac{96 g oxygen x 350g Al (C2H3O2)3}{204 gAl (C2H3O2)3} [/tex] of oxygen
= 164.5 grams
Thus 164.5 grams of oxygen is present in 350 g of Al(C₂H₃O₂)₃
Answer:
The option B is a correct answer.
Explanation:
Mass of aluminum ethoxide = 350 g
Moles of aluminum ethoxide = [tex]\frac{350 g}{204 g/mol}=1.7157 mol[/tex]
In 1 mole of aluminum ethoxide there are 6 moles of oxygen atom.
Then 1.7157 moles of aluminum ethoxide will have :
[tex]1.7157 mol\times 6 =10.294 mol[/tex]
Mass of 10.291 mole of oxygen atom = 10.294 mol 16 g/mol = 164.70 g[/tex]
The closest option to our answer is option B.
Hence, option B is correct answer.
