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

The grams of aluminum sulfide formed are ; 12.51 g

Explanation:

Using Formula,

Calculate the moles of Al and S :

Given  mass of Al = 9 g

Molar mass of Al = 27 g/ mole

Mole of Al = 0.33

Given  mass of S =  8 g

Molar mass of S = 32 g/mole

Mole of S  = 0.25

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

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

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 :

= 0.0833 moles of Al2S3

= 0.083 mole

Molar mass of

= 150.16 g/mole

= 12.51 g

She is not correct because ∠1 is an alternate exterior angle with ∠3, not the angle that measures 105°. So m∠1 = 75°.

Explanation:

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