Question

Aluminum metal reacts with chlorine gas to form aluminum chloride. The reaction can be represented by the following balanced chemical equation:

2 Al(s) + 3 Cl2(g) → 2 AlCl3(s).


If you were trying to react 52.9 grams of chlorine gas completely, how many grams of aluminum would you need?


_____________________ [grams of Aluminum] Do NOT enter the unit in your answer. Report your answer with 3 SFs.

Answer 1

Solution :-

Given to us is the Reaction of Aluminium metal with Chlorine gas . That is ,

`\boxed{\red{\bf \underset{\blue{Aluminium}}{2Al(s)}+\underset{\blue{Chlorine}}{3Cl_2(g)}\longrightarrow \underset{\blue{Aluminium\: chloride}}{2AlCl_3}}}`

Now Mass of 1 mole aluminium = 27g . So mass of 2 moles of it would be 27×2 g = 54g .

Also , Mass of 1 mole of chlorine gas = 35.5 g . So mass of 2 Atoms will be 71g . And mass of 3 moles of atom will be 71 × 3g = 213g .

Now , use unitary method ;

213 g of Chlorine reacts with 54g of Aluminium.

1 g of Chlorine will react with
`\sf(54g)/(213g)` of Aluminium.

Hence 52.9 g of Chlorine will react with
`\sf (54)/(213)*52.9g`=
`\bf 13.41g` of Aluminium.