Question

Calculate the maximum numbers of moles and grams of H2S that can form when 153.0 g of aluminum sulfide reacts with 143.0 g of water: Al2S3 + H2O → Al(OH)3 + H2S [unbalanced]

Answer 1

Answer:

We'll form 3.057 moles H2S , that is 104.2 grams

Explanation:

Step 1: Data given

Mass of Al2S3 = 153.0 grams

Mass if water = 143.0 grams

Molar mass Al2S3 = 150.16 g/mol

Molar mass H2O = 18.02 g/mol

Step 2: The balanced equation

Al2S3 + 6H2O → 2Al(OH)3 + 3H2S

Step 3: Calculate moles Al2S3

Moles Al2S3 = mass Al2S3 / molar mass Al2S3

Moles Al2S3 = 153.0 grams / 150.16 g/mol

Moles Al2S3 = 1.019 moles

Step 4: Calculate moles H2O

Moles H2O = mass H2O / molar mass H2O

Moles H2O = 143.0 grams / 18.02 g/mol

Moles H2O = 7.936 moles

Step 5: Calculate limiting reactant

For 1 mol Al2S3 we need 6 moles H2O to produce 2 moles Al(OH)3 and 3 moles H2S

Al2S3 is the limiting reactant. It will completely be consumed (1.019 moles).

H2O is in excess. There will react 6*1.019 = 6.114 moles

There will remain 7.936 - 6.114 = 1.822 moles

Step 6: Calculate moles H2S

For 1 mol Al2S3 we need 6 moles H2O to produce 2 moles Al(OH)3 and 3 moles H2S

For 1.019 moles we'll have 3*1.019 = 3.057 moles

Step 7: Calculate mass H2S

Mass H2S = moles H2S * molar mass H2S

Mass H2S = 3.057 moles * 34.1 g/mol

Mass H2S = 104.2 grams

We'll form 3.057 moles H2S , that is 104.2 grams

Answer 2

The maximum number of moles is 3.057 moles H₂S  , that is 104.2 grams.

Given:

Mass of Al₂S₃ = 153.0 grams

Mass of H₂O = 143.0 grams

Molar mass Al₂S₃ = 150.16 g/mol

Molar mass H₂O = 18.02 g/mol

Balanced chemical reaction:

Al₂S₃ + 6H₂O → 2Al(OH)₃ + 3H₂S

Calculation for number of moles:

$\text{Moles of }Al_2S_3 = \frac{\text{ Mass of } Al_2S_3}{\text{Molar mass of }Al_2S_3}\\\\\text{Moles of }Al_2S_3 = \frac{153.0 grams}{150.16 g/mol} \\\\\text{Moles of }Al_2S_3= 1.019 moles$

$\text{ Moles of } H_2O =\frac{\text{ Mass of } H_2O}{\text {Molar mass of } H_2O} \\\\\text{ Moles of } H_2O = \frac{143.0 grams}{18.02 g/mol} \\\\\text{ Moles of } H_2O = 7.936 moles$

Determination of Limiting reagent:

For 1 mol Al₂S₃ we need 6 moles H₂O  to produce 2 moles Al(OH)₃ and 3 moles H₂S

Al₂S₃ is the limiting reactant. It will completely be consumed (1.019 moles).

H₂O  is in excess. There will react 6*1.019 = 6.114 moles

There will remain 7.936 - 6.114 = 1.822 moles

Calculation for moles:

For 1 mol Al₂S₃ we need 6 moles H₂O to produce 2 moles Al(OH)₃ and 3 moles H₂S

For 1.019 moles we'll have 3*1.019 = 3.057 moles

Calculation for mass of H₂S :

Mass  of H₂S = moles H₂S  * molar mass H₂S

Mass of H₂S  = 3.057 moles * 34.1 g/mol

Mass of H₂S  = 104.2 grams

Thus, It will form 3.057 moles H₂S  , that is 104.2 grams.

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