How many oxygen atoms are present in 2.74 g of Al_2(SO_4)_3?

• A. 1.16 × 10^{22} oxygen atoms
• B. 2.48 × 10^{21} oxygen atoms
• C. 8.07 × 10^{20} oxygen atoms
• D. 4.32 × 10^{23} oxygen atoms

Answer: (A)

Verified step by step guidance

1. Calculate the molar mass of aluminum sulfate, Al\_2(SO\_4)\_3, by summing the atomic masses of all atoms in the formula: 2 aluminum (Al), 3 sulfur (S), and 12 oxygen (O) atoms. Use the atomic masses: Al = 26.98 g/mol, S = 32.07 g/mol, O = 16.00 g/mol.
2. Determine the number of moles of Al\_2(SO\_4)\_3 in 2.74 g by using the formula: \(\text{moles} = \frac{\text{mass}}{\text{molar mass}}\).
3. Calculate the total number of moles of oxygen atoms by multiplying the moles of Al\_2(SO\_4)\_3 by 12, since there are 12 oxygen atoms per formula unit.
4. Use Avogadro's number, \$6.022 \times 10^{23}\( atoms/mol, to convert moles of oxygen atoms to the actual number of oxygen atoms: \)\text{number of atoms} = \text{moles of oxygen} \times 6.022 \times 10^{23}$.
5. Interpret the result to find the number of oxygen atoms present in 2.74 g of Al\_2(SO\_4)\_3 and compare it with the given options.