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

2Al (s) + 3Cl2 (g) →2AlCl3 (s)

You are given 10.0 g of aluminum and 15.0 g of chlorine gas. If you had excess chlorine, how many moles of of aluminum chloride could be produced from 10.0 g of aluminum?

0.3706 moles of aluminum chloride would be produced from 10.0 g of aluminum

Explanation:

chemical equation:

2Al (s) + 3Cl2 (g) → 2AlCl3 (s)

Given dа ta:

mass of aluminium = 10 g

mass of chlorine = 15 g

First of all we will calculate the moles of each reactant:

moles of aluminium = 10 g / 26.98 g/mol

moles of aluminium = o. 3706 mol

moles of Cl2 = 15 g / 10.90 g/mol

moles of Cl2 = 0.2116 mol

now we compare the moles aluminium with aluminium chloride

Al : AlCl3

2 : 2

1 : 1

0.3706 : 0.3706

now we compare the moles chlorine with aluminium chloride

Cl : AlCl3

3 : 2

0.2116 : 2/3 * 0.2116 = 0.1412 mole

although chlorine is limiting reactant but in question we are given excess chlorine so 10 g of aluminium will produce 0.3706 moles of aluminium chloride.

mass of AlCl3 = 0.3706 mol * 133.34 g/mol

mass of AlCl3 = 50.13 g