Question

how many grams of aluminum sulfide can form from the reaction 9.00 g of aluminum with 8.00 g of sulfur

Answer 1

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}=(9.00g)/(27g/mole)=0.333moles`

`\text{ Moles of }S_8=\frac{\text{ Mass of }S_8}{\text{ Molar mass of }S_8}=(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
`(0.0312)/(3)* 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
`(0.0312)/(3)* 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* \text{ Molar mass of }Al_2S_3`

`\text{ Mass of }Al_2S_3=(0.0832moles)* (150.2g/mole)=12.5g`

Therefore, the mass of aluminum sulfide form from the reaction can be 12.5 grams.

Answer 2

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