Final answer:
To determine the number of grams of oxygen in 28.5 grams of aluminum nitrate, calculate the molar mass of Al(NO3)3 and find the ratio of oxygen to aluminum nitrate.
Step-by-step explanation:
To determine the number of grams of oxygen in 28.5 grams of aluminum nitrate, we need to calculate the molar mass of Al(NO3)3. The molar mass of Al is 26.98 g/mol, N is 14.01 g/mol, and O is 16.00 g/mol.
The molar mass of Al(NO3)3 is:
26.98 g/mol + (3 x 14.01 g/mol) + (9 x 16.00 g/mol) = 213.00 g/mol
Next, we need to find the ratio of oxygen to aluminum nitrate:
(16.00 g/mol ÷ 213.00 g/mol) x 28.5 g = 2.15 g
Therefore, there would be approximately 2.15 grams of oxygen in 28.5 grams of aluminum nitrate.