Question

Aluminum reacts with chlorine gas to form aluminum chloride via the following reaction: 2Al(s)+3Cl₂g)→2AlCl₃((s) You are given 21.0 g of aluminum and 26.0 g of chlorine gas. Part A If you had excess chlorine, how many moles of of aluminum chloride could be produced from 21.0 g of aluminum? Express your answer to three significant figures and include the appropriate units.

Answer 1

To determine the moles of aluminum chloride produced from 21.0 g of aluminum, first calculate the moles of aluminum using its molar mass. the balanced equation to convert the moles of aluminum to moles of aluminum chloride. Finally, convert the moles of aluminum chloride to grams using its molar mass.

Step-by-step explanation:

To determine how many moles of aluminum chloride can be produced from 21.0 g of aluminum, we need to use the balanced equation and the molar masses of aluminum and aluminum chloride. The molar mass of aluminum is 26.98 g/mol and the molar mass of aluminum chloride is 133.34 g/mol. First, we calculate the number of moles of aluminum by dividing its mass by its molar mass: Moles of Al = 21.0 g ÷ 26.98 g/mol = 0.778 mol.

According to the balanced equation, 2 moles of aluminum react to form 2 moles of aluminum chloride. Therefore, 0.778 moles of aluminum will produce 0.778 moles of aluminum chloride: Moles of AlCl₃ = 0.778 mol × 2 mol AlCl₃/2 mol Al = 0.778 mol Finally, we convert the moles of aluminum chloride to grams by multiplying it by its molar mass: Mass of AlCl₃ = 0.778 mol × 133.34 g/mol ≈ 103.5 g.