Aluminum reacts with chlorine gas to form aluminum chloride according to the reaction below. 2 Al(s)
+
3 Cl2 (g)
2 AlCl3 (s)
How many moles of aluminum chloride can be produced from 2.12 L of chlorine gas at a
temperature of 24 °C and a pressure of 2.27 atm?

0.131mol

Explanation:

Firstly, we need to find the number of moles of chlorine gas (Cl2) by using the formula:

PV = nRT

Where;

P = pressure (atm)

V = volume (L)

n = number of moles (mol)

R = gas constant (0.0821 Latm/molK)

T = temperature (K)

Using PV = nRT

n of Chlorine = PV/RT

Based on the information provided in this question, V = 2.12 L, P = 2.27atm, T = 24°C = 24 + 273 = 297K

n = 2.27 × 2.12/ 0.0821 × 297

n = 4.8124/24.3837

n = 0.197

number of moles of Chlorine gas = 0.197mol

The balanced chemical equation provided in this question is as follows;

2 Al(s) + 3Cl2 (g) → 2AlCl3 (s)

3 moles of Cl2 produce 2 moles of AlCl3

0.197moles of Cl2 will produce;

0.197 × 2/3

= 0.394/3

= 0.131mol of AlCl3.