Under the right conditions aluminum will react with chlorine to produce aluminum chloride. 2 Al + 3 Cl2 - 2 AICI3
How many grams of aluminum are needed to react completely with 11.727 liters of chlorine?
CORRECT ANSWER IS 9.42, but what are the steps to get that answer ?

9.42 g

Explanation:

Step 1: Write the balanced equation

2 Al + 3 Cl₂ ⇒ 2 AICI₃

Step 2: Calculate the moles corresponding to 11.727 L of Cl₂

Since the conditions are not specified, we will assume Cl₂ is at standard temperature and pressure. At STP, 1 mole of Cl₂ occupies 22.4 L.

11.727 L × 1 mol/22.4 L = 0.524 mol

Step 3: Calculate the moles of Al needed to react with 0.524 moles of Cl₂

The molar ratio of Al to Cl₂ is 2:3. The moles of Al needed are 2/3 × 0.524 mol = 0.349 mol

Step 4: Calculate the mass corresponding to 0.349 moles of Al

The molar mass of Al is 26.98 g/mol.

0.349 mol × 26.98 g/mol = 9.42 g