1. **State the problem:** Find the number of moles in 98.3 g of Aluminum Hydroxide, Al(OH)$_3$. 2. **Formula used:** Number of moles $n = \frac{\text{mass}}{\text{molar mass}}$ 3. **Calculate molar mass of Al(OH)$_3$:** - Atomic mass of Al = 26.98 g/mol - Atomic mass of O = 16.00 g/mol - Atomic mass of H = 1.008 g/mol Molar mass = $26.98 + 3 \times (16.00 + 1.008) = 26.98 + 3 \times 17.008 = 26.98 + 51.024 = 77.994$ g/mol 4. **Calculate moles:** $$n = \frac{98.3}{77.994}$$ 5. **Simplify fraction:** $$n = \frac{\cancel{98.3}}{\cancel{77.994}} \approx 1.26$$ 6. **Answer:** There are approximately **1.26 moles** of Aluminum Hydroxide in 98.3 g.