Question

If you measured out 28.5 grams of aluminum nitrate, how many grams of oxygen would there be?

a. 28.94
b. -44.00
c. 48.00

Answer 1

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.