Question

Be sure to answer all parts. The balanced equation for the reaction of aluminum metal and chlorine gas is 2Al(s) + 3Cl2(g) → 2AlCl3(s) Assume that 0.80 g Al is mixed with 0.23 g Cl2. (a) What is the limiting reactant? Cl2 Al (b) What is the maximum amount of AlCl3, in grams, that can be produced? g AlCl3

Answer 1

Answer: a)
`Cl_2` is the limiting reagent

b) 0.27 g of
`AlCl_3` will be produced.

Step-by-step explanation:

To calculate the moles :

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

`\text{Moles of} Al=(0.80g)/(27g/mol)=0.030moles`

`\text{Moles of} Cl_2=(0.23g)/(71g/mol)=0.003moles`

`2Al(s)+3Cl_2(g)\rightarrow 2AlCl_3(s)`

According to stoichiometry :

3 moles of
`Cl_2` require = 2 moles of
`Al`

Thus 0.003 moles of
`Cl_2` will require=
`(2)/(3)* 0.003=0.002moles` of
`Al`

Thus
`Cl_2` is the limiting reagent as it limits the formation of product and
`Al` is the excess reagent.

b) As 3 moles of
`Cl_2` give = 2 moles of
`AlCl_3`

Thus 0.003 moles of
`Cl_2` give =
`(2)/(3)* 0.003=0.002moles` of
`AlCl_3`

Mass of
`AlCl_3=moles* {\text {Molar mass}}=0.002moles* 133g/mol=0.27g`

Thus 0.27 g of
`AlCl_3` will be produced.