Answer to Question #349672 in General Chemistry for jasmine

Question

If 8.28 L of hydrogen gas at SATP is produced, how many moles of aluminium were reacted?

Answer 1

under normal conditions occupies a volume of 22.4 l per 1 mol gas. so we find how many moles are in 8.28 l of H₂ :

n (H₂) = V / 22,4 = 8,28 / 22,4 = 0,37 mol

2Al + 6HCl = 2AlCl₃ + 3H₂

2mol(54 g) Al —> 3mol H₂

X g Al —> 0,37 mol H₂

X = 54 * 0,37 / 3 = 6,66 g

Answer: 6,66 g Al