How many oxygen atoms are in 6.70 g of Al2(SO4)3? Express your answer using scientific notation with two decimal places.
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To determine the number of oxygen atoms in 6.70 g of Al2(SO4)3, you need to use the concept of molar mass and Avogadro's number.
1. First, calculate the molar mass of Al2(SO4)3:
- Aluminum (Al) has a molar mass of 26.98 g/mol.
- Sulfur (S) has a molar mass of 32.07 g/mol.
- Oxygen (O) has a molar mass of 16.00 g/mol.
Al2(SO4)3 consists of 2 aluminum atoms, 3 sulfur atoms, and 12 oxygen atoms.
Molar mass of Al2(SO4)3 = (2 * molar mass of Al) + (3 * molar mass of S) + (12 * molar mass of O)
Molar mass of Al2(SO4)3 = (2 * 26.98 g/mol) + (3 * 32.07 g/mol) + (12 * 16.00 g/mol)
Molar mass of Al2(SO4)3 = 342.15 g/mol
2. Use the molar mass calculated in step 1 to find the number of moles of Al2(SO4)3:
Moles = Mass / Molar mass
Moles of Al2(SO4)3 = 6.70 g / 342.15 g/mol
Moles of Al2(SO4)3 ≈ 0.0196 mol
3. Finally, multiply the number of moles of Al2(SO4)3 by the number of oxygen atoms per molecule:
Number of oxygen atoms = Moles of Al2(SO4)3 * 12 (since there are 12 oxygen atoms in one molecule of Al2(SO4)3)
Number of oxygen atoms ≈ 0.0196 mol * 12
Number of oxygen atoms ≈ 0.2347 mol
Expressing the answer in scientific notation with two decimal places, you would have:
Number of oxygen atoms = 2.35 x 10^22 (rounded to two decimal places)