If 0.50 mol of Al reacts completely with excess H2SO4, how many moles of Al2(SO4)3 can be formed?

• A. 0.75 mol
• B. 0.25 mol
• C. 0.50 mol
• D. 1.00 mol

Answer: (B)

Verified step by step guidance

1. Write the balanced chemical equation for the reaction between aluminum (Al) and sulfuric acid (H\_2SO\_4). The reaction produces aluminum sulfate (Al\_2(SO\_4)\_3) and hydrogen gas (H\_2): \[2Al + 3H_2SO_4 \rightarrow Al_2(SO_4)_3 + 3H_2\]
2. Identify the mole ratio between aluminum and aluminum sulfate from the balanced equation. According to the equation, 2 moles of Al produce 1 mole of Al\_2(SO\_4)\_3.
3. Use the given amount of aluminum (0.50 mol) and apply the mole ratio to find the moles of aluminum sulfate formed. Set up the proportion: \[\text{moles of } Al_2(SO_4)_3 = 0.50 \text{ mol Al} \times \frac{1 \text{ mol } Al_2(SO_4)_3}{2 \text{ mol Al}}\]
4. Calculate the moles of aluminum sulfate formed by performing the multiplication in the proportion (do not compute the final number here, just set up the expression).
5. Interpret the result as the amount of aluminum sulfate formed when 0.50 mol of aluminum reacts completely with excess sulfuric acid.