Aluminum metal reacts with hydrochloric acid according to the following equation: 2 A l left parenthesis s right parenthesis plus 6 H C l left parenthesis a q right parenthesis rightwards arrow 2 A l C l subscript 3 left parenthesis a q right parenthesis plus 3 H subscript 2 left parenthesis g right parenthesis

If 28.1 grams of aluminum reacts with 46 mL of a 1 M HCl solution, how many grams of AlCl3 will be generated? Hint: You must determine the limiting reactant

18 g

Explanation:

Step 1: Write the balanced equation

2 Al + 6 HCl ⇒ 2 AlCl₃ + 3 H₂

Step 2: Calculate the moles of Al

The molar mass of Al is 26.98 g/mol.

28.1 g × 1 mol/26.98 g = 1.04 mol

Step 3: Calculate the moles of HCl

46 mL of 1 M HCl react.

0.046 L × 1 mol/L = 0.046 mol

Step 4: Determine the limiting reactant

The theoretical molar ratio of Al to HCl is 2:6 = 0.33:1.

The real molar ratio of Al to HCl is 1.04:0.046 = 22.6:1

According to this, the limiting reactant is HCl.

Step 5: Calculate the mass of AlCl₃ generated from 0.046 moles of HCl

The molar ratio of HCl to AlCl₃ is 6:2. The molar mass of AlCl₃ is 133.34 g/mol.

0.046 mol HCl × 6 mol HCl/2 mol AlCl₃ × 133.34 g AlCl₃/1 mol AlCl₃ = 18 g AlCl₃

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