The mass of aluminum chloride formed is approximately 158.62 g when 30.0 mL of 1.00 M HCl reacts with 32 g of powdered aluminum.
When hydrochloric acid (HCl) reacts with aluminum (Al), it forms aluminum chloride (AlCl₃) and hydrogen gas (H₂). The balanced chemical equation for this reaction is:
2 Al + 6 HCl → 2 AlCl₃ + 3 H₂
To determine the mass of aluminum chloride formed, we need to consider the limiting reactant. First, we find the moles of each reactant. The molar mass of aluminum is approximately 27 g/mol. Thus, 32 g of aluminum is about 32 g / 27 g/mol ≈ 1.19 moles of aluminum.
Given that the concentration of HCl is 1.00 M, the moles of HCl are 1.00 mol/L * 0.030 L = 0.030 moles.
Considering the balanced equation, the stoichiometric ratio between aluminum and HCl is 2:6. Therefore, 1.19 moles of aluminum would react with (6/2) * 1.19 = 3.57 moles of HCl. Since we have less HCl than required, it is the limiting reactant.
Now, we find the moles of aluminum chloride formed. For every 2 moles of aluminum, 2 moles of aluminum chloride are produced. Thus, the moles of aluminum chloride formed would be (2/2) * 1.19 = 1.19 moles.
Finally, we calculate the mass of aluminum chloride using its molar mass (133.34 g/mol):
Mass of AlCl₃ = 1.19 mol * 133.34 g/mol ≈ 158.62 g
The question probable may be:
What is the mass of aluminum chloride formed when a student adds 30.0 mL of 1.00 M HCl to 32 g of powdered aluminum?