Question

If you had excess chlorine, how many moles of of aluminum chloride could be produced from 12.0 g of aluminum? express your answer to three significant figures and include the appropriate units.

Answer 1

1) If we had excess Cl2, then limiting reagent is Al

so moles of Al will be the same as the moles of AlCl3 produced at the output

moles of Al = 23/27 = 0.8518 moles

2)If we had excess Al , then limiting reagent is Cl2

so moles of Cl2 will be the 3/2 times as the moles of AlCl3 produced at the output

moles of Cl2 = 28/71 = 0.394 moles

hence moles of AlCl3 = 0.394*2/3 = 0.2629 moles

Answer 2

Answer : The number of moles of aluminium chloride will be 0.444 moles.

Solution : Given,

Mass of Al = 12.0 g

Molar mass of Al = 27 g/mole

First we have to calculate the moles of Al.

`\text{ Moles of }Al=\frac{\text{ Mass of }Al}{\text{ Molar mass of }Al}=(12.0g)/(27g/mole)=0.444moles`

Now we have to calculate the moles of
`AlCl_3`

The balanced chemical reaction is,

`2Al+3Cl_2\rightarrow 2AlCl_3`

From the balanced reaction we conclude that

As, 2 mole of
`Al` react to give 2 mole of
`AlCl_3`

So, 0.444 moles of
`Al` react to give 0.444 moles of
`AlCl_3`

Therefore, the number of moles of aluminium chloride will be 0.444 moles.