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Sagot :
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
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
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 [tex]S_8[/tex] = 8.00 g
Molar mass of Al = 27 g/mole
Molar mass of [tex]S_8[/tex] = 256 g/mole
Molar mass of [tex]Al_2S_3[/tex] = 150.2 g/mole
First we have to calculate the moles of Al and [tex]S_8[/tex].
[tex]\text{ Moles of }Al=\frac{\text{ Mass of }Al}{\text{ Molar mass of }Al}=\frac{9.00g}{27g/mole}=0.333moles[/tex]
[tex]\text{ Moles of }S_8=\frac{\text{ Mass of }S_8}{\text{ Molar mass of }S_8}=\frac{8.00g}{256g/mole}=0.0312moles[/tex]
Now we have to calculate the limiting and excess reagent.
The balanced chemical reaction is,
[tex]16Al(s)+3S_8(s)\rightarrow 8Al_2S_3(s)[/tex]
From the balanced reaction we conclude that
As, 3 mole of [tex]S_8[/tex] react with 16 mole of [tex]Al[/tex]
So, 0.0312 moles of [tex]S_8[/tex] react with [tex]\frac{0.0312}{3}\times 16=0.166[/tex] moles of [tex]Al[/tex]
From this we conclude that, [tex]Al[/tex] is an excess reagent because the given moles are greater than the required moles and [tex]S_8[/tex] is a limiting reagent and it limits the formation of product.
Now we have to calculate the moles of [tex]Al_2S_3[/tex]
From the reaction, we conclude that
As, 3 mole of [tex]S_8[/tex] react to give 8 mole of [tex]Al_2S_3[/tex]
So, 0.0312 moles of [tex]S_8[/tex] react to give [tex]\frac{0.0312}{3}\times 8=0.0832[/tex] moles of [tex]Al_2S_3[/tex]
Now we have to calculate the mass of [tex]Al_2S_3[/tex]
[tex]\text{ Mass of }Al_2S_3=\text{ Moles of }Al_2S_3\times \text{ Molar mass of }Al_2S_3[/tex]
[tex]\text{ Mass of }Al_2S_3=(0.0832moles)\times (150.2g/mole)=12.5g[/tex]
Therefore, the mass of aluminum sulfide form from the reaction can be 12.5 grams.
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