Question

15.00 g of aluminum sulfide (150.1 g/mol) and 10.00 g of water (18.02 g/mol) react until the limiting reactant is used up. Calculate the mass of H2S (34.08 g/mol) that can be produced from these reactants. Notice that you will need to balance the reaction equation.
___Al2S3(s)+ ___H2O > ___Al(OH)3(s)+ ___H2S(g)

a. 13.89 g
b. 10.21 g
c. 19.67 gd. 9.456 g
e. 1.108 g

Answer 1

Answer:

9.457 g H₂S

Explanation:

Al₂S₃ + 6H₂O => 2Al(OH)₃ + 3 H₂S

moles Al₂S₃ = 15g/150.1 g/mol = 0.0999 mole / 1 => 0.0999

moles H₂O = 10 g/18.02 g/mol = 0.555 mole / 6 => 0.0.0925 => LR*

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*A quick way to ID the LR is to convert given mass data to moles and divide each reactant by its respective coefficient value. The smaller resulting value of the set is the limiting reactant. However, one must work problem using the moles calculated in previous step.

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Moles H₂S produced = 3/6(0.555 mole) = 0.2775 mole H₂S x 34.08 g/mol = 9.457 g H₂S.

Answer 2

Answer:

The mass of H2S produced is 9.456 grams (option D is correct)

Explanation:

Step 1: Data given

Mass of aluminium sulfide (Al2S3) = 15.00 grams

Molar mass of Al2S3 = 150.1 g/mol

Mass of H2O = 10.0 grams

Molar mass of H2O = 18.02 g/mol

Molar mass of H2S = 34.08 g/mol

Step 2: The balanced equation

Al2S3(s)+ 6H2O → 2Al(OH)3(s)+ 3H2S(g)

Step 3: Calculate moles Al2S3

Moles Al2S3 = mass Al2S3 / molar mass Al2S3

Moles Al2S3 = 15.00 grams / 150.1 g/mol

Moles Al2S3 = 0.100 moles

Step 4: Calculate moles H2O

Moles H2O = 10.0 grams / 18.02 g/mol

Moles H2O = 0.555 moles

Step 5: Calculate the limiting reactant

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

H2O is the limiting reactant. It will completely be consumed (0.555 moles). Al2S3 is in excess. There will react 0.555 / 6 = 0.0925 moles

There will remain 0.100 - 0.0925 = 0.0075 moles

Step 6: Calculate moles of H2S

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

For 0.555 moles H2O we'll have 0.555/ 2 = 0.2775 moles H2S

Step 7: Calculate mass H2S

Mass H2S = 0.2775 moles * 34.08 g/mol

Mass H2S = 9.456 grams

The mass of H2S produced is 9.456 grams (option D is correct)