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