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

The mass of hydrogen sulfide (H₂S) that could be produced is 9.456 g. The correct option is d. 9.456 g

First,

We will write a balanced chemical equation for the reaction

The balanced chemical equation for the reaction is

Al₂S₃(s) + 6H₂O(l) → 2Al(OH)₃(s) + 3H₂S(g)

This means,

1 mole of aluminum sulfide reacts with 6 moles of water to produce 2 moles of  aluminum hydroxide and 3 moles of hydrogen sulfide

Now, we will determine the number of moles of each reactant present

Mass = 15.00 g

Molar mass = 150.1 g/mol

From the formula

∴ Number of moles of Al₂S₃ =

Number of moles of Al₂S₃ = 0.0999 moles

Mass = 10.00 g

Molar mass = 18.02 g/mol

∴ Number of moles of water =

Number of moles of water = 0.554939 moles

From the balanced chemical equation

1 mole of aluminum sulfide reacts with 6 moles of water

Then,

x moles of aluminum sulfide will react with 0.554939 moles of water

x =

x = 0.09248

∴ The number of moles of aluminum sulfide that reacted is 0.09248 moles

Since

1 mole of aluminum sulfide reacts with 6 moles of water to produce 3 moles of hydrogen sulfide

Then,

0.09248 moles of aluminum sulfide reacts with 0.554939 moles of water to produce 0.2774695 moles of hydrogen sulfide

∴ The number of moles of hydrogen sulfide produced in the reaction is 0.2774695 moles

Now, for the mass of hydrogen sulfide (H₂S) produced

Molar mass of H₂S = 34.08 g/mol

From the formula

Mass = Number of moles ×  Molar mass

∴ Mass of H₂S produced = 0.2774695 × 34.08

Mass of H₂S produced = 9.456 g

Hence, the mass of hydrogen sulfide (H₂S) that could be produced is 9.456 g. The correct option is d. 9.456 g

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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)