Question

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

2 Al + 3 Cl2 → 2 AlCl3

How many grams of aluminum chloride could be produced from 34.00 g of aluminum and 39.00 g of chlorine gas? How much of the excess reactant will be left after the reaction has gone to completion? (Al = 26.98 amu, Cl = 35.45 amu)

Mass of product = Blank 1 grams of AlCl3

Mass of excess reactant remaining = Blank 2 grams of excess reactant left unreacted

Answer 1

1)

Molar mass of Al = 26.98 g/mol

mass of Al = 34.0 g

we have below equation to be used:

number of mol of Al,

n = mass of Al/molar mass of Al

=(34.0 g)/(26.98 g/mol)

= 1.26 mol

Molar mass of Cl2 = 70.9 g/mol

mass of Cl2 = 39.0 g

we have below equation to be used:

number of mol of Cl2,

n = mass of Cl2/molar mass of Cl2

=(39.0 g)/(70.9 g/mol)

= 0.5501 mol

we have the Balanced chemical equation as:

2 Al + 3 Cl2 ---> 2 AlCl3

2 mol of Al reacts with 3 mol of Cl2

for 1.26 mol of Al, 1.89 mol of Cl2 is required

But we have 0.5501 mol of Cl2

so, Cl2 is limiting reagent

we will use Cl2 in further calculation

Molar mass of AlCl3 = 1*MM(Al) + 3*MM(Cl)

= 1*26.98 + 3*35.45

= 133.33 g/mol

From balanced chemical reaction, we see that

when 3 mol of Cl2 reacts, 2 mol of AlCl3 is formed

mol of AlCl3 formed = (2/3)* moles of Cl2

= (2/3)*0.5501

= 0.3667 mol

we have below equation to be used:

mass of AlCl3 = number of mol * molar mass

= 0.3667*1.333*10^2

= 48.89 g

Answer: mass of profuct = 48.9 g

2)

From balanced chemical reaction, we see that

when 3 mol of Cl2 reacts, 2 mol of Al reacts

mol of Al reacted = (2/3)* moles of Cl2

= (2/3)*0.5501

= 0.3667 mol

mol of Al remaining = mol initially present - mol reacted

mol of Al remaining = 1.26 - 0.3667

mol of Al remaining = 0.8935 mol

Molar mass of Al = 26.98 g/mol

we have below equation to be used:

mass of Al,

m = number of mol * molar mass

= 0.8935 mol * 26.98 g/mol

= 24.11 g

Answer: Mass of excess reactant remaining = 24.1 g Al left unreacted