Aluminum and chlorine gas react together to make Aluminum chloride. What is the limiting reactant if 20.0g of Al and 30.0g of Cl2 are used?how much aluminum chloride was formed?

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To determine the limiting reactant, we need to compare the amount of each reactant used to their respective stoichiometric coefficients in the balanced chemical equation.

The balanced chemical equation for the reaction between aluminum (Al) and chlorine gas (Cl2) to form aluminum chloride (AlCl3) is as follows:

2Al + 3Cl2 -> 2AlCl3

First, we need to determine the number of moles for each reactant using their given masses and molar masses:

Molar mass of Al = 26.98 g/mol
Molar mass of Cl2 = 70.90 g/mol

Number of moles of Al = (mass of Al) / (molar mass of Al)
Number of moles of Cl2 = (mass of Cl2) / (molar mass of Cl2)

Number of moles of Al = 20.0 g / 26.98 g/mol ≈ 0.741 mol
Number of moles of Cl2 = 30.0 g / 70.90 g/mol ≈ 0.423 mol

Now, we compare the moles of each reactant to their respective stoichiometric coefficients in the balanced chemical equation:

Al : Cl2 = 0.741 mol : 0.423 mol

Since the stoichiometric ratio of Al to Cl2 is 2:3, we can see that there is an excess of chlorine gas, as there is more Cl2 than what is required to react with all the available aluminum.

To determine the limiting reactant, we compare the number of moles of each reactant to their respective stoichiometric ratios:

Al:Cl2 (from the balanced equation) = 2:3

In this case, the number of moles of Al is limiting the reaction, as there are fewer moles of Al compared to the required ratio. Hence, aluminum is the limiting reactant.

To calculate the mass of aluminum chloride formed, we use the stoichiometry from the balanced equation. From the equation, we see that 2 moles of Al react to form 2 moles of AlCl3. Therefore, the stoichiometric ratio of Al to AlCl3 is 2:2 or 1:1.

Since the number of moles of AlCl3 formed is the same as the number of moles of Al, we can simply use the moles of Al to calculate the mass of AlCl3:

Number of moles of AlCl3 = Number of moles of Al = 0.741 mol

Now, to find the mass of AlCl3 formed, we use the equation:

Mass of AlCl3 = (Number of moles of AlCl3) * (molar mass of AlCl3)

Molar mass of AlCl3 = (molar mass of Al) + 3 * (molar mass of Cl)

Molar mass of AlCl3 = 26.98 g/mol + 3 * (35.45 g/mol) ≈ 135.33 g/mol

Mass of AlCl3 = 0.741 mol * 135.33 g/mol ≈ 100.21 g

Therefore, approximately 100.21 grams of aluminum chloride were formed.