Question

What volume (in mL) of 0.250 M HCI would be

required to completely react with 4.10 g of Al in the
following chemical reaction?
2 Al(s) + 6 HCl(aq) → 2 AICI3 (aq) + 3 H₂(g)

Answer 1

Answer: 1820 mL (to 3 sf)

Step-by-step explanation:

The atomic mass of aluminum is 26.9815385 g/mol, so 4.10 g of aluminum is equal to 4.10/26.9815385 = 0.15195575300497 moles of Al.

From the coefficients of the equation, we know that for every 2 moles of aluminum consumed, 6 moles of HCl are consumed.

So, this means we need 0.15195575300497(6/2) = 0.45586725901491 moles of HCl.

Substituting into the molarity formula,

• 0.250 = 0.45586725901491/(liters of HCl)
• liters of HCl = 0.45586725901491/0.250
• liters of HCl = 1.8234690360596 L = 1820 mL (to 3 sf)