Question

PLEASE HELP!!!!!!!

25.0g of Aluminum metal is combined with
45.Og Copper(ll) Chloride to produce Aluminum Chloride
and Copper metal
1. Write and balance the chemical equation.
2. Write the needed mole ratio between reactants.
3. What is the mole ratio you have from the data?
4. What is in excess and what is limited? (Show Work)
5. What is the theoretical yield of Aluminum Chloride?

Answer 1

Answer: 1.
`2Al+3CuCl_2\rightarrow 2AlCl_3+3Cu`

2. 3 moles of
`CuCl_2` : 2 moles of
`Al`

3. 0.33 moles of
`CuCl_2` : 0.92 moles of
`Al`

4.
`CuCl_2` is the limiting reagent and
`Al` is the excess reagent.

5. Theoretical yield of
`AlCl_3` is 29.3 g

Step-by-step explanation:

To calculate the moles :

`\text{Moles of solute}=\frac{\text{given mass}}{\text{Molar Mass}}`

`\text{Moles of} Al=(25.0g)/(27g/mol)=0.92moles`

`\text{Moles of} CuCl_2=(45.0g)/(134g/mol)=0.33moles`

The balanced chemical equation is:

`2Al+3CuCl_2\rightarrow 2AlCl_3+3Cu`

According to stoichiometry :

3 moles of
`CuCl_2` require = 2 moles of
`Al`

Thus 0.33 moles of
`CuCl_2` will require=
`(2)/(3)* 0.33=0.22moles` of
`Al`

Thus
`CuCl_2` is the limiting reagent as it limits the formation of product and
`Al` is the excess reagent.

As 3 moles of
`CuCl_2` give = 2 moles of
`AlCl_3`

Thus 0.33 moles of
`CuCl_2` give =
`(2)/(3)* 0.33=0.22moles` of
`AlCl_3`

Theoretical yield of
`AlCl_3=moles* {\text {Molar mass}}=0.22moles* 133.34g/mol=29.3`

Thus 29.3 g of aluminium chloride is formed.