How many grams of aluminum (Al) must react with excess sulfuric acid (H_2SO_4) to produce 2.6 L of hydrogen gas (H_2) at STP?

• A. 8.1 g
• B. 6.0 g
• C. 4.7 g
• D. 2.6 g

Answer: (B)

Verified step by step guidance

1. Write the balanced chemical equation for the reaction between aluminum and sulfuric acid: 2Al + 3H_2SO_4 \rightarrow Al_2(SO_4)_3 + 3H_2.
2. Identify the volume of hydrogen gas produced (2.6 L) and recall that at STP (Standard Temperature and Pressure), 1 mole of any gas occupies 22.4 L.
3. Calculate the moles of hydrogen gas produced using the formula: moles H_2 = \frac{\text{volume of } H_2}{22.4 \text{ L/mol}} = \frac{2.6}{22.4}.
4. Use the mole ratio from the balanced equation to find moles of aluminum needed: from 3 moles of H_2, 2 moles of Al react, so moles Al = \frac{2}{3} \times \text{moles H}_2.
5. Calculate the mass of aluminum required by multiplying moles of Al by its molar mass (approximately 27 g/mol): mass Al = moles Al \times 27 \text{ g/mol}.