Stoichiometric calculation of aluminum mass from hydrogen volume Steps:

• Total H₂ volume is 143 cm³; moles H₂ = 143 / 24,000 = 0.00596 mol (using 24,000 cm³/mol).
• From 2Al + 6HCl → 2AlCl₃ + 3H₂, moles Al = (2/3) × 0.00596 mol = 0.00397 mol.
• Mass Al = 0.00397 mol × 27 g/mol = 0.107 g.
• This matches the fixed mass added, as reaction goes to completion with excess Al.

Why A is correct:

• It uses the stoichiometric ratio and ideal gas molar volume to find the exact mass from total product gas.

Why the others are wrong:

• B: Mass is a one-time addition, not a rate (g/min); data shows fixed amount, not continuous feeding.
• C: Highest rate is (131 - 53) cm³ / 2 min = 39 cm³/min between 6–8 min.
• D: Initial rate is slowest (high acid concentration), but accelerates later, likely due to Al oxide layer dissolving.

Final answer: A