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Molar mass of Al = 26.98 g/mol

mass of Al = 33.0 g

we have below equation to be used:

number of mol of Al,

n = mass of Al/molar mass of Al

=(33.0 g)/(26.98 g/mol)

= 1.223 mol

Molar mass of Cl2 = 70.9 g/mol

mass of Cl2 = 38.0 g

we have below equation to be used:

number of mol of Cl2,

n = mass of Cl2/molar mass of Cl2

=(38.0 g)/(70.9 g/mol)

Part B

we have the Balanced chemical equation as:

2 Al + 3 Cl2 ---> 2 AlCl3

2 mol of Al reacts with 3 mol of Cl2

for 1.2231 mol of Al, 1.8347 mol of Cl2 is required

But we have 0.536 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)*0.536

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explanation:

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