How many liters of chlorine are needed to produce 5.64 moles of aluminum chloride? Need help ASAP

190 L

Explanation:

Given that:

Number of moles = mass/molar mass

number of moles = 5.64

molar mass of aluminum chloride = 133.34 g/mol

∴

The balanced equation is:

2 Al + 3 Cl2 --> 2AlCl3.

From above reaction:

5.64 mole of  AlCl3 reacts with 3 mole Cl2  to give 2 mole AlCl3

i.e

= 8.46 mole Cl2 gas.

At Standard temperature and pressure:

= (22.4 L/1 mole gas) * (8.46 mole Cl2)

= 190 L chlorine gas