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

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

Answer 2

Answer:

17.22 g s the amount of $Al_2S_3$ remains.

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:

$moles = \frac{Mass\ taken}{Molar\ mass}$

Thus,

$Moles= \frac{20.00\ g}{150.17\ g/mol}$

$Moles_{Al_2S_3}= 0.1332\ mol$

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:

$moles = \frac{Mass\ taken}{Molar\ mass}$

Thus,

$Moles= \frac{2.00\ g}{18.02\ g/mol}$

$Moles_{H_2O}= 0.1110\ mol$

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 $\frac{1}{6}\times 0.1110$ 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.