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]

Question

Chemistry

Tristian Ellis

27 September, 06:01

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]

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Answers (2)

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Danny Downs · Answer 1

1.02 moles of H₂S can be produced in the reaction which means a mass of 33.7 g

Explanation:

We determine the balance reaction:

Al₂S₃ + 6H₂O → 2Al (OH) ₃ + 3H₂S

We have the mass of each reactant, therefore we can determine the limiting reactant. First of all we convert the mass to moles:

153 g / 150.14 g/mol = 1.02 moles of sulfide

143 g / 18 g/mol = 7.94 moles of water

6 moles of water need 1 mol of sulfide to react

Then, 7.94 moles of water may need (7.94.1) / 6 = 1.32 moles

I only have 1.02 moles, so the Al₂S₃ is my limiting reactant.

1 mol of sulfide can produce 1 mol of hydrogen sulfide (Ratio is 1:1)

Then, 1.02 moles of sulfide must produce 1.02 moles of H₂S

If we convert the moles to mass → 1.02 mol. 33.06 g/1mol = 33.7 g

Margaret Chase · Answer 2

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