Question

20.1 g of aluminum and 219 g of chlorine gas react until all of the aluminum metal has been converted to AlCl3. The balanced equation for the reaction is the following.

2 Al(s) + 3 Cl2(g) → 2 AlCl3(s)
What is the quantity of chlorine gas left, in grams, after the reaction has occurred, assuming that the reaction goes to completion? (The formula mass of aluminum metal, Al, is 26.98 g/mol, and the formula mass of chlorine gas, Cl2, is 70.90 g/mol.)

Answer 1

The amount of Cl2 gas left , after the reaction goes to completion is : 139.655 grams

Step-by-step explanation:

Molar mass : It is the mass in grams present in one mole of the substance.

Moles of the substance is calculated by:

`Moles=(Mass)/(Molar\ mass)`

`2Al(s)+3Cl_(2)(g)\leftarrow 2AlCl_(3)(g)`

According to this equation:

2 mole of Al = 3 mole of Cl2 = 2 mole of AlCl3

Molar mass of Al = 27.0 g/mol

Mass of Al = 20.1 gram

Moles of Al present in the reaction :

`Moles=(Mass)/(Molar\ mass)`

`Moles=(20.1)/(26.98)`

Moles of Al = 0.744

Similarly calculate the moles of Cl2

Molar mass of Cl2 = 71.0 g/mol

Mass = 219 gram

`Moles=(Mass)/(Molar\ mass)`

`Moles=(219)/(70.98)`

Moles of Cl2 = 3.08 moles

According to equation,

2 mole of Al reacts with = 3 mole of Cl2

1 moles of Al reacts with = 3/2 mole of Cl2

0.744 moles of Al reacts with = 3/2(0.744) moles of Cl2

= 1.116 moles of Cl2

But actually present Cl2 = 3.08 moles

Hence Al is the limiting reagent , and Cl2 is the excess reagent.

The whole Aluminium Al get consumed during the reaction.

The amount of Cl2 in excess = Total Cl2 - Cl2 consumed

Cl2 in excess = 3.08 - 1.116 = 1.964 moles

Cl2 in grams = 1.964 x 70.90 = 139.655 grams