Question

Create the Equation: How many grams of Aluminum Chloride would be made from 42.7 L of Chlorine gas at STP reacting with 50.0 g of Aluminum?

Answer 1

Answer: 169.3 g of
`AlCl_3` will be produced

Step-by-step explanation:

To calculate the moles, we use the equation:

`\text{Number of moles}=\frac{\text{Given mass}}{\text {Molar mass}}`

moles of
`Al`

`\text{Number of moles}=(50.0g)/(27g/mol)=1.85moles`

moles of
`Cl_2`

According to ideal gas equation:

`PV=nRT`

P = pressure of gas = 1 atm (STP)

V = Volume of gas = 42.7 L

n = number of moles = ?

R = gas constant =
`0.0821Latm/Kmol`

T =temperature =
`273K` (at STP)

`n=(PV)/(RT)`

`n=(1atm* 42.7L)/(0.0820 L atm/K mol* 273K)=1.90moles`

`2Al+3Cl_2\rightarrow 2AlCl_3`

According to stoichiometry:

3 moles of
`Cl_2` reacts with = 2 moles of aluminium

Thus 1.90 moles of
`Cl_2`reacts with=
`(2)/(3)* 1.90=1.27` moles of aluminium

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

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

Thus 1.90 moles of
`Cl_2` give =
`(2)/(3)* 1.90=1.27moles` of
`AlCl_3`

Mass of
`AlCl_3=moles* {\text {Molar mass}}=1.27moles* 133.34g/mol=169.3g`

Thus 169.3 g of
`AlCl_3` will be produced from the given masses of both reactants.