what is the theoretical yield of KOH + Al --> Al(OH3) + H2
http://www.newton.dep.anl.gov/askasci/chem03/chem03682.htm
Theoretical yield refers to the maximum amount of product that can be obtained from a chemical reaction, assuming 100% efficiency and complete conversion of all reactants.
To calculate the theoretical yield for the reaction:
1. Start by writing a balanced chemical equation:
2KOH + 2Al → 2KAlO2 + H2
Here, 2 moles of KOH react with 2 moles of Al to produce 2 moles of KAlO2 and 1 mole of H2.
2. Determine the molar mass of the reactants and products:
- Molar mass of KOH (potassium hydroxide) = 56.11 g/mol
- Molar mass of Al (aluminum) = 26.98 g/mol
- Molar mass of KAlO2 (potassium aluminate) = 96.14 g/mol
- Molar mass of H2 (hydrogen gas) = 2.016 g/mol
3. Convert the given mass of KOH to moles:
- Let's assume we have 10 grams of KOH.
- Convert grams to moles using the molar mass of KOH:
10 g KOH * (1 mol KOH / 56.11 g KOH) = 0.178 moles KOH
4. Use stoichiometry to determine the moles of Al(OH)3 and H2 produced:
- From the balanced equation, the stoichiometric ratio between KOH and Al(OH)3 is 1:1.
- Therefore, the moles of Al(OH)3 produced will also be 0.178 moles.
- From the balanced equation, the stoichiometric ratio between KOH and H2 is 2:1.
- Therefore, half the moles of KOH will give the moles of H2 produced:
0.178 moles KOH * (1 mole H2 / 2 moles KOH) = 0.089 moles H2
5. Convert the moles of Al(OH)3 and H2 to grams:
- Al(OH)3:
0.178 moles * (96.14 g/mol) = 17.14 grams
- H2:
0.089 moles * (2.016 g/mol) = 0.179 grams
Therefore, the theoretical yield of Al(OH)3 is 17.14 grams, and the theoretical yield of H2 is 0.179 grams.