What how can I calculate the amount in moles and the mass in grams of KOH needed to convert 1.50 g of aluminum to KAl(OH)4. Balanced equation = 2 Al (s) + 2 KOH (aq) + 6 H2O (l)->2 KAl(OH)4 (aq) + 3 H2 (g)
Please show steps ..
To calculate the amount in moles and the mass in grams of KOH needed to convert 1.50 g of aluminum to KAl(OH)4, we will use stoichiometry.
1. Calculate the molar mass of KOH:
- The molar mass of K is 39.10 g/mol.
- The molar mass of O is 16.00 g/mol.
- The molar mass of H is 1.01 g/mol.
- Therefore, the molar mass of KOH is (39.10 + 16.00 + 1.01) g/mol = 56.11 g/mol.
2. Convert the given mass of aluminum to moles:
- The molar mass of aluminum (Al) is 26.98 g/mol.
- To convert grams to moles, divide the mass by the molar mass:
1.50 g Al / 26.98 g/mol = 0.0556 mol Al.
3. Determine the mole ratios between Al and KOH from the balanced equation:
- From the balanced equation, we can see that 2 mol of Al reacts with 2 mol of KOH.
- Therefore, the mole ratio is 2 mol Al : 2 mol KOH.
4. Calculate the moles of KOH required:
- Since the mole ratio tells us that 2 mol of KOH is required for every 2 mol of Al, we can conclude that 0.0556 mol Al will require 0.0556 mol KOH.
5. Convert the moles of KOH to grams:
- To convert moles to grams, multiply the moles by the molar mass of KOH:
0.0556 mol KOH x 56.11 g/mol = 3.11 g KOH.
Therefore, you would need 0.0556 moles of KOH (which is approximately 3.11 grams) to convert 1.50 g of aluminum to KAl(OH)4.
To calculate the amount in moles and the mass in grams of KOH needed to convert 1.50 g of aluminum to KAl(OH)4, we need to follow a few steps.
Step 1: Determine the molar mass of the substances involved.
- The molar mass of Al (Aluminum) = 26.98 g/mol
- The molar mass of KOH (Potassium Hydroxide) = 56.11 g/mol
- The molar mass of KAl(OH)4 (Potassium Tetrahydroxoaluminate) = 258.32 g/mol
Step 2: Convert the given mass of aluminum to moles.
- Use the formula: Moles = Mass / Molar mass
- Moles of Aluminum = 1.50 g / 26.98 g/mol
Step 3: Determine the stoichiometric ratio between aluminum and KOH.
- From the balanced equation: 2Al (s) + 2KOH (aq) + 6H2O (l) -> 2KAl(OH)4 (aq) + 3H2 (g)
- The stoichiometric ratio of Al to KOH is 1:1. Therefore, the moles of aluminum will be equal to the moles of KOH.
Step 4: Calculate the moles and mass of KOH.
- Moles of KOH = Moles of Aluminum = 1.50 g / 26.98 g/mol
- Mass of KOH = Moles of KOH × Molar mass of KOH
Let's calculate the values:
Step 1: Molar mass of substances:
- Molar mass of Al = 26.98 g/mol
- Molar mass of KOH = 56.11 g/mol
- Molar mass of KAl(OH)4 = 258.32 g/mol
Step 2: Moles of aluminum:
- Moles of Aluminum = 1.50 g / 26.98 g/mol = 0.0556 mol
Step 3: Stoichiometric ratio:
- The stoichiometric ratio of Al to KOH is 1:1.
Step 4: Moles and mass of KOH:
- Moles of KOH = Moles of Aluminum = 0.0556 mol
- Mass of KOH = Moles of KOH × Molar mass of KOH = 0.0556 mol × 56.11 g/mol = 3.10 g
Therefore, to convert 1.50 g of aluminum to KAl(OH)4, you would need 0.0556 mol of KOH, which is equivalent to 3.10 grams of KOH.
mols Al = grams/atomic mass
Use the coefficients in the balanced equation to convert mols Al to mols KOH.
Then convert mols KOH to g. g = mols x molar mass.
All of these are done this same way with these three steps. (Sometimes the first step is mols = M x L if it is a soln. Sometimes they stick a percent yield at the end. But basically these three steps will do it.)