Question

In an experiment, 9 g of aluminum react with 8 g of sulfur to form aluminum sulfide. Calculate the grams of aluminum sulfide formed. Al + S → Al2S3

Answer 1

The grams of aluminum sulfide formed are ; 12.51 g

Step-by-step explanation:

Using Formula,

`mole = (given\ mass)/(Molar\ mass)`

Calculate the moles of Al and S :

Given mass of Al = 9 g

Molar mass of Al = 27 g/ mole

`mole = (9)/(27)`

Mole of Al = 0.33

Given mass of S = 8 g

Molar mass of S = 32 g/mole

`mole = (8)/(32)`

Mole of S = 0.25

The balanced chemical equation for the reaction between aluminium and sulfur is :

`2Al(s) + 3S(s)\rightarrow Al_(2)S_(3)(s)`

Here ,

2 mole of Al needs 3 moles of S

1 mole of Al needs 3/2(= 1.5) moles of S

hence ,

0.33 mole of Al should require

`(3)/(2)* 0.33` moles of S

Sulfur needed = 0.495 mole

Available S = 0.25 mole

So there is less sulfur than required , S is the limiting reagent

Amount of S decide the Amount of Al2S3 formed

3 mole of S produce 1 mole of Al2S3

1 mole of S produce 1/3 mole of Al2S3

0.25 mole will produce :

`0.25*(1)/(3)`

= 0.0833 moles of Al2S3

`Al_(2)S_(3)` = 0.083 mole

Molar mass of

`Al_(2)S_(3)` = 150.16 g/mole

`mole = (given\ mass)/(Molar\ mass)`

`given\ mass = moles* molar\ mass`

`given\ mass = 0.083* 150.16`

= 12.51 g