Question

What is the amount of Al2S3 remains when 20.00 g of Al2S3 and 2.00 g of H2O are reacted? A few of the molar masses are as follows: Al2S3 = 150.17 g/mol, H2O = 18.02 g/mol and Al2S3(s) + 6 H2O(l) → 2 Al(OH)3(s) + 3 H2S(g)

Answer 1

17.22 g s the amount of
`Al_2S_3` remains.

Step-by-step explanation:

Moles of
`Al_2S_3`:-

Mass = 20.00 g

Molar mass of
`Al_2S_3` = 150.17 g/mol

The formula for the calculation of moles is shown below:

`moles = (Mass\ taken)/(Molar\ mass)`

Thus,

`Moles= (20.00\ g)/(150.17\ g/mol)`

`Moles_(Al_2S_3)= 0.1332\ mol`

Moles of
`H_2O`:-

Mass = 2.00 g

Molar mass of
`H_2O` = 18.02 g/mol

The formula for the calculation of moles is shown below:

`moles = (Mass\ taken)/(Molar\ mass)`

Thus,

`Moles= (2.00\ g)/(18.02\ g/mol)`

`Moles_(H_2O)= 0.1110\ mol`

According the given reaction:-

`Al_2S_3_((s)) + 6 H_2O_((l))\rightarrow 2 Al(OH)_3_((s)) + 3 H_2S_((g))`

1 mole of
`Al_2S_3` reacts with 6 moles of
`H_2O`

0.1332 mole of
`Al_2S_3` reacts with 0.1332*6 moles of
`H_2O`

Moles of
`H_2O` required = 0.7992 mol

Available moles of
`H_2O` = 0.1110 mol

Limiting reagent is the one which is present in small amount. Thus,
`H_2O` is limiting reagent.

The formation of the product is governed by the limiting reagent. So,

6 moles of
`H_2O` reacts with 1 mole of
`Al_2S_3`

Also,

1 mole of
`H_2O` reacts with 1/6 mole of
`Al_2S_3`

0.1110 mole of
`H_2O` reacts with
`(1)/(6)* 0.1110` mole of
`Al_2S_3`

Moles of
`Al_2S_3` reacted = 0.0185 moles

Thus, moles of
`Al_2S_3` unreacted = 0.1332 moles - 0.0185 moles = 0.1147 moles

Moles of
`Al_2S_3` unreacted = 0.1147 moles

Mass = Moles*Molar mass = 0.1147moles*150.17 g/mol = 17.22 g

17.22 g s the amount of
`Al_2S_3` remains.

Answer 2

From the equation, we can tell that 1 mol of Al₂S₃ requires 6 moles of water.
The molar ratio is 1/6
Moles of Al₂S₃ present = 20/150.17
= 0.133
Moles of water present = 2/18.02
= 0.111
The moles of Al₂S₃ that will react are:
0.111/6
= 0.0185
The remaining amount:
0.133 - 0.0185
= 0.1145 mol
Or
0.1145 * 150.17
= 17.19 grams