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Answer :
Final answer:
In 97.8 moles of Alâ‚‚(SOâ‚„)₃, there are about 7.07 × 10²âµ oxygen atoms. This is calculated by multiplying the number of moles by Avogadro's number and the number of oxygen atoms per formula unit of Alâ‚‚(SOâ‚„)₃.
Explanation:
To answer the question of how many atoms of O are in 97.8 moles of Al₂(SO₄)₃, it's necessary to first understand the formula Al₂(SO₄)₃. This formula tells us that for each formula unit of Al₂(SO₄)₃, there are two Al atoms, three SO₄ groups, and each SO₄ group contains four O atoms.
To calculate the number of O atoms in the formula unit, you multiply the number of SO₄ groups (3) by the number of O atoms in each group (4). This results in 12 O atoms per formula unit of Al₂(SO₄)₃.
Now, knowing that one mole contains 6.022 × 10²³ entities (Avogadro's number), you calculate the total number of O atoms in 97.8 moles by multiplying the number of moles by Avogadro's number, and then by the number of O atoms in each formula unit. So, 97.8 moles x 6.022 x 10²³ entities/mole x 12 O atoms/entity gives you about 7.07 × 10²âµ O atoms.
To determine the number of atoms of Oxygen (O) in 97.8 moles of Al₂(SO₄)₃, we need to find the number of moles of O in one mole of Al₂(SO₄)₃, and then multiply it by 97.8.
The formula for Al₂(SO₄)₃ indicates that it contains 12 O atoms in one mole of the compound. Therefore, in one mole of Al₂(SO₄)₃, there are 12 moles of O atoms.
Multiplying 12 moles of O by 97.8 moles of Al₂(SO₄)₃ gives us 1173.6 moles of O atoms.
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