Aluminum reacts with chlorine gas to form aluminum chloride via the following reaction: 2al(s)+3cl2(> 2alcl3(s) you are given 18.0g of aluminum and 23.0g of chlorine gas. part a: if you had excess chlorine, how many moles of of aluminum chloride could be produced from 18.0g of aluminum? express your answer numerically in moles. part b: if you had excess aluminum, how many moles of aluminum chloride could be produced from 23.0g of chlorine gas, cl2? express your answer numerically in moles.

(a) Moles of Aluminium chloride formed = 0.6671 moles.

(b) Moles of Aluminium chloride formed = 0.2163 moles.

Explanation:

Limiting reagent is the one which is present in small molar amount. It got exhausted at the end of the reaction and the formation of the products is governed by it.

(a) Aluminium is limiting.

The formula for the calculation of moles is shown below:

Given: For Aluminium

Given mass = 18.0 g

Molar mass of Aluminium = 26.982 g/mol

Moles of Aluminium = 18.0 g / 26.982 g/mol = 0.6671 moles

According to the reaction,

2 moles of aluminium on reaction forms 2 moles of aluminum chloride.

Thus,

0.6671 moles of aluminium on reaction forms 0.6671 moles of aluminum chloride.

Moles of Aluminium chloride formed = 0.6671 moles.

(b) Chlorine gas is limiting.

The formula for the calculation of moles is shown below:

Given: For chlorine gas

Given mass = 23.0 g

Molar mass of chlorine gas = 70.906 g/mol

Moles of chlorine gas = 23.0 g / 70.906 g/mol = 0.3244 moles

According to the reaction,

3 moles of chlorine gas on reaction forms 2 moles of aluminum chloride.

1 moles of chlorine gas on reaction forms 2/3 moles of aluminum chloride.

Thus,

0.3244 moles of chlorine gas on reaction forms 2/3 * 0.3244 moles of aluminum chloride.

Moles of Aluminium chloride formed = 0.2163 moles.

given:

mass of na = 115 g

excess cl2

to determine:

mass of nacl produced

explanation:

given reaction is-

2na(s) + cl2(g) → 2nacl(s)

since cl2 is in excess, na will be the limiting reagent

as per the reaction stoichiometry na: nacl = 1: 1

i.e. moles of na reacted = moles of nacl formed

now, # moles of na = mass of na/atomic mass

                                  = 115 g/23 g.mol-1 = 5 moles

therefore, moles of nacl = 5

molar mass of nacl = 58 g/mol

mass of nacl = 5 moles * 58 g.mol-1 = 290 g

ans: amount of nacl produced = 290 g