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How many moles are in 98.3 g of [tex]$Al(OH)_3$[/tex]?

Answer :

To determine the moles in 98.3 g of [tex]$Al(OH)_3$[/tex], follow these steps:

1. Calculate the molar mass of [tex]$Al(OH)_3$[/tex]. The compound contains:
- 1 atom of Aluminum (Al)
- 3 hydroxide groups, each containing 1 Oxygen (O) and 1 Hydrogen (H)

2. Use the approximate atomic masses:
- Atomic mass of Al: 26.98 g/mol
- Atomic mass of O: 16.00 g/mol
- Atomic mass of H: 1.01 g/mol

3. First, calculate the molar mass contribution of one hydroxide group, [tex]$OH$[/tex]:
[tex]$$
M(OH) = 16.00 \, \text{g/mol} + 1.01 \, \text{g/mol} = 17.01 \, \text{g/mol}
$$[/tex]

4. Since there are 3 hydroxide groups in [tex]$Al(OH)_3$[/tex], their total mass is:
[tex]$$
3 \times 17.01 \, \text{g/mol} = 51.03 \, \text{g/mol}
$$[/tex]

5. Now, add the mass of one Al atom:
[tex]$$
M(Al(OH)_3) = 26.98 \, \text{g/mol} + 51.03 \, \text{g/mol} = 78.01 \, \text{g/mol}
$$[/tex]

6. Finally, calculate the number of moles by using the formula:
[tex]$$
\text{Number of moles} = \frac{\text{Mass}}{\text{Molar mass}} = \frac{98.3 \, \text{g}}{78.01 \, \text{g/mol}} \approx 1.26009 \, \text{mol}
$$[/tex]

Thus, there are approximately [tex]$1.26009$[/tex] moles in 98.3 g of [tex]$Al(OH)_3$[/tex].

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