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

Answer:

The grams of aluminum sulfide formed are ; 12.51 g

Explanation:

Using Formula,

$mole = \frac{given\ mass}{Molar\ mass}$

Calculate the moles of Al and S :

Given  mass of Al = 9 g

Molar mass of Al = 27 g/ mole

$mole = \frac{9}{27}$

Mole of Al = 0.33

Given  mass of S =  8 g

Molar mass of S = 32 g/mole

$mole = \frac{8}{32}$

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

$\frac{3}{2}\times 0.33$ 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 :

$0.25\times\frac{1}{3}$

= 0.0833 moles of Al2S3

$Al_{2}S_{3}$ = 0.083 mole

Molar mass of

$Al_{2}S_{3}$ = 150.16 g/mole

$mole = \frac{given\ mass}{Molar\ mass}$

$given\ mass = moles\times molar\ mass$

$given\ mass = 0.083\times 150.16$

= 12.51 g