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)

the only liquid elements at standard temperature and pressure are bromine (br) and mercury (hg). although, elements caesium (cs), rubidium (rb), francium (fr) and gallium (ga) become liquid at or just above room temperature.

explanation:

c.

explanation: