* Chemical Reactions are Necessary: The conversion of aluminum sulfide to aluminum hydroxide requires a chemical reaction. You need to know the specific reaction that will be used. For example, you might react aluminum sulfide with water or an acid.
* Stoichiometry: Even knowing the reaction, you'll need to use stoichiometry to calculate the mass of aluminum hydroxide produced. Stoichiometry is the study of the quantitative relationships between reactants and products in chemical reactions. It uses balanced chemical equations to determine the amounts of substances involved.
To get an answer, you need:
1. The Balanced Chemical Equation: This will tell you the mole ratio between aluminum sulfide and aluminum hydroxide.
2. The Molar Masses: You need the molar masses of aluminum sulfide (Al₂S₃) and aluminum hydroxide (Al(OH)₃) to convert between grams and moles.
Let's illustrate with an example reaction:
Reaction:
Al₂S₃ + 6H₂O → 2Al(OH)₃ + 3H₂S
Steps to Calculate the Mass of Aluminum Hydroxide:
1. Balance the Equation: The equation above is already balanced.
2. Molar Masses:
* Al₂S₃: (2 x 26.98 g/mol Al) + (3 x 32.06 g/mol S) = 150.16 g/mol
* Al(OH)₃: (26.98 g/mol Al) + (3 x 16.00 g/mol O) + (3 x 1.01 g/mol H) = 78.00 g/mol
3. Convert Grams of Aluminum Sulfide to Moles:
* moles Al₂S₃ = (17.2 g Al₂S₃) / (150.16 g/mol Al₂S₃) = 0.115 mol Al₂S₃
4. Use the Mole Ratio from the Balanced Equation:
* The balanced equation shows that 1 mol of Al₂S₃ produces 2 moles of Al(OH)₃.
* moles Al(OH)₃ = (0.115 mol Al₂S₃) x (2 mol Al(OH)₃ / 1 mol Al₂S₃) = 0.230 mol Al(OH)₃
5. Convert Moles of Aluminum Hydroxide to Grams:
* grams Al(OH)₃ = (0.230 mol Al(OH)₃) x (78.00 g/mol Al(OH)₃) = 17.94 g Al(OH)₃
Therefore, if you react 17.2 g of aluminum sulfide with water according to the given reaction, you would obtain approximately 17.94 g of aluminum hydroxide.
