Respuesta :
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
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:
[tex]\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[/tex]
[tex]\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[/tex]
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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