Answer to Question #140029 in General Chemistry for Khaled

Question

Aluminum reacts with chlorine gas to form aluminum chloride via the following reaction:

2Al(s)+3Cl2(g)→2AlCl3(s)

You are given 34.0 g of aluminum and 39.0 g of chlorine gas.

1-If you had excess chlorine, how many moles of aluminum chloride could be produced from 34.0 g of aluminum?
Express your answer to three significant figures and include the appropriate units.
2-If you had excess aluminum, how many moles of aluminum chloride could be produced from 39.0 g of chlorine gas, Cl2?
Express your answer to three significant figures and include the appropriate units.

Answer 1

Q140029

Aluminum reacts with chlorine gas to form aluminum chloride via the following reaction.

2Al (s) + 3Cl₂ (g) --> 2AlCl₃ (s)

You are given 34.0g of aluminum and 39.0g of chlorine gas.

1-If you had excess chlorine, how many moles of aluminum chloride could be produced from 34.0 g of aluminum?

Express your answer to three significant figures and include the appropriate units.

2-If you had excess aluminum, how many moles of aluminum chloride could be produced from 39.0 g of chlorine gas, Cl2?

Express your answer to three significant figures and include the appropriate units.

Solution :

Part 1) The given reaction is

2Al (s) + 3Cl₂ (g) --> 2AlCl₃ (s)

mass of aluminum we are given = 34.0g , and Cl₂ is in excess.

So amount of aluminum chloride produced will depend on the mass of aluminum.

Atomic mass of Al = 26.982g/mol

moles of Al = 34.0g Al * 1mol Al / 26.982g Al = 1.2601 mol of Al

Next we use the mole ratio of Al and AlCl₃ from the reaction and find the ‘moles of AlCl₃ ‘ .

mol ratio of Al and AlCl₃ in the reaction = 2 : 2 = 1 : 1

So, moles of AlCl₃ formed = 1.2601 mol of Al * 1 mol AlCl₃ /1 mol Al = 1.2601mol of AlCl₃

In question we are told to write the answer in 3 significant figure.

So, moles of Aluminum chloride formed from given mass of Aluminum = 1.26 mol of AlCl₃

Part 2) In second part, we are given mass of chlorine gas, Cl₂ = 39.0g and Aluminum is in excess.

We will convert 39.0g of Cl₂ to moles using molar mass of Cl₂ .

Atomic mass of Cl = 35.453g/mol

Molar mass of Cl₂ = 2 * atomic mass of Cl = 2 * 35.453g/mol = 70.906g/mol

moles of Cl₂ = 39.0g Cl₂ * 1 mol Cl₂ /70.906 g of Cl₂ = 0.5500 mol of Cl₂ .

Next we will use the mol ratio of Cl₂ and AlCl₃ from the reaction and find the ‘moles of AlCl₃ ‘ formed from 0.5500 mol of Cl₂

2Al (s) + 3Cl₂ (g) --> 2AlCl₃ (s)

mol ratio of Cl₂ and AlCl₃ in the given reaction is 3 : 2

moles of AlCl₃ formed = 0.5500 mol Cl₂ * 2 mol AlCl₃ /3 mol Cl₂ = 0.3667 mol AlCl₃

which in 3 significant figure = 0.367 mol AlCl₃ .

Hence, moles of Aluminum chloride formed from given mass of chlorine gas is 0.367mol