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

Question

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

2Al(s)+3Cl2(g)?2AlCl3(s)

You are given 19.0g of aluminum and 24.0g of chlorine gas.

1.If you had excess chlorine, how many moles of of aluminum chloride could be produced from 19.0g 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 24.0g of chlorine gas, Cl2?

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

Explanation / Answer

1)
molar mass of Al = 27.0 g/mol
number of mol of Al = (given mass)/(molar mass)
= 19.0/27.0
= 0.704 mol

reaction
2Al(s)   +   3Cl2(g) -- >    2AlCl3(s)

when we have Cl2 is in excess
then, Al act as limiting reagent
according to reaction
2 mol of Al give 2 mol of AlCl3
1 mol of Al give 1 mol of AlCl3
0.704 mol of Al give 0.704 mol of AlCl3
so,
number of mol of AlCl3 formed = 0.704 mol

2)
molar mass of Cl2 = 70.8 g/mol
number of mol of Cl2 = (given mass)/(molar mass)
= 24.0/70.8
= 0.339 mol

reaction
2Al(s)   +   3Cl2(g) -- >    2AlCl3(s)

when we have Al is in excess
then, Cl2 act as limiting reagent
according to reaction
3 mol of Cl2 give 2 mol of AlCl3
1 mol of Cl2 give 2/3 mol of AlCl3
0.339 mol of Cl2 give (2/3)*0.339 mol of AlCl3
0.339 mol of Cl2 give 0.226 mol of AlCl3
so,
number of mol of Cl2 formed = 0.226 mol

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