How many grams of aluminum sulfide can form from the reaction of 9.00 g of aluminum with 8.00 g of sulfur?
Answer (3)
Answer : The mass of aluminum sulfide form from the reaction can be 12.5 grams.
Solution : Given,
Mass of Al = 9.00 g
Mass of S 8 = 8.00 g
Molar mass of Al = 27 g/mole
Molar mass of S 8 = 256 g/mole
Molar mass of A l 2 S 3 = 150.2 g/mole
First we have to calculate the moles of Al and S 8 .
Moles of A l = Molar mass of A l Mass of A l = 27 g / m o l e 9.00 g = 0.333 m o l es
Moles of S 8 = Molar mass of S 8 Mass of S 8 = 256 g / m o l e 8.00 g = 0.0312 m o l es
Now we have to calculate the limiting and excess reagent.
The balanced chemical reaction is,
16 A l ( s ) + 3 S 8 ( s ) → 8 A l 2 S 3 ( s )
From the balanced reaction we conclude that
As, 3 mole of S 8 react with 16 mole of A l
So, 0.0312 moles of S 8 react with 3 0.0312 × 16 = 0.166 moles of A l
From this we conclude that, A l is an excess reagent because the given moles are greater than the required moles and S 8 is a limiting reagent and it limits the formation of product.
Now we have to calculate the moles of A l 2 S 3
From the reaction, we conclude that
As, 3 mole of S 8 react to give 8 mole of A l 2 S 3
So, 0.0312 moles of S 8 react to give 3 0.0312 × 8 = 0.0832 moles of A l 2 S 3
Now we have to calculate the mass of A l 2 S 3
Mass of A l 2 S 3 = Moles of A l 2 S 3 × Molar mass of A l 2 S 3
Mass of A l 2 S 3 = ( 0.0832 m o l es ) × ( 150.2 g / m o l e ) = 12.5 g
Therefore, the mass of aluminum sulfide form from the reaction can be 12.5 grams.
2Al + 3S ---> Al₂S₃
1 mole of Al = 27g 1 mole of S = 32g 1 mole of Al₂S₃ = 150g
according to the reaction: 2 27g Al ------------------ 3 32g S 9g Al---------------------------- x g S x = 16g S >> s, alluminium is excess
according to the reaction: 3*32g S----------------------- 150g Al₂S₃ 8g S------------------------------ x Al₂S₃ x = 12,5g Al₂S₃
The reaction of 9.00 g of aluminum with 8.00 g of sulfur produces 12.5 grams of aluminum sulfide. Sulfur is the limiting reactant in this reaction. We calculate the required moles and convert them to mass to find the answer.
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