How many grams of hydrogen are released by the reaction of 1.00 g of aluminum with excess potassium hydroxide solution assuming 100% yield
Well, isn't it electrifying to talk about chemical reactions? Now, let me do some calculations with a punchline.
The balanced equation for the reaction between aluminum and potassium hydroxide is:
2Al + 2KOH + 6H2O -> 2KAl(OH)4 + 3H2
According to the equation, 2 moles of aluminum will produce 3 moles of hydrogen gas. Now, let's calculate the moles of aluminum first.
The molar mass of aluminum (Al) is approximately 26.98 g/mol. So, 1.00 gram of aluminum is equal to 1.00 g / 26.98 g/mol = 0.037 mol of aluminum.
Since the reaction is assumed to have 100% yield, 0.037 mol of aluminum will produce 3 moles of hydrogen gas.
Now, converting moles of hydrogen gas to grams. The molar mass of hydrogen (H2) is approximately 2.02 g/mol. So, 3 moles of hydrogen gas is equal to 3 mol x 2.02 g/mol = 6.06 grams of hydrogen.
Therefore, 1.00 gram of aluminum will release 6.06 grams of hydrogen gas. Tada!
To determine the number of grams of hydrogen released, we need to use the balanced chemical equation for the reaction between aluminum and potassium hydroxide. The balanced equation is as follows:
2 Al (s) + 2 KOH (aq) + 6 H₂O (l) → 2 KAl(OH)₄ (aq) + 3 H₂ (g)
From the balanced equation, we can see that 2 moles of aluminum react to produce 3 moles of hydrogen gas. To calculate the amount of hydrogen, we need to determine the moles of aluminum and then use the stoichiometry of the reaction.
Step 1: Convert grams of aluminum to moles:
The molar mass of aluminum (Al) is 26.98 g/mol. Therefore,
1.00 g Al * (1 mol Al / 26.98 g Al) = 0.03705 mol Al
Step 2: Use the stoichiometry of the reaction to determine the moles of hydrogen produced:
From the balanced equation, we know that for every 2 moles of aluminum, 3 moles of hydrogen are produced. Therefore,
0.03705 mol Al * (3 mol H₂ / 2 mol Al) = 0.05558 mol H₂
Step 3: Convert moles of hydrogen to grams:
The molar mass of hydrogen (H₂) is 2.02 g/mol. So,
0.05558 mol H₂ * (2.02 g H₂ / 1 mol H₂) = 0.1122 g H₂
Therefore, approximately 0.1122 grams of hydrogen are released in this reaction assuming 100% yield.
To determine the amount of hydrogen gas released in the reaction between aluminum and potassium hydroxide, we need to use stoichiometry and molar mass of the substances involved.
First, let's write the balanced chemical equation for the reaction:
2Al + 6KOH -> 3H2 + 2K3AlO3
From the balanced equation, we can see that 2 moles of aluminum react with 6 moles of potassium hydroxide to produce 3 moles of hydrogen gas.
Next, we need to find the molar mass of aluminum. The molar mass of aluminum is 26.98 g/mol.
Now, we can set up the calculation:
1. Convert the mass of aluminum to moles:
moles of aluminum = mass (Al) / molar mass (Al)
moles of aluminum = 1.00 g / 26.98 g/mol
2. Use the mole ratio from the balanced equation to determine the number of moles of hydrogen gas produced:
moles of hydrogen = moles of aluminum x (3 moles H2 / 2 moles Al)
3. Convert the moles of hydrogen to grams:
mass of hydrogen = moles of hydrogen x molar mass of hydrogen
The molar mass of hydrogen is 2.02 g/mol.
Therefore, the calculation for the grams of hydrogen gas released can be done as follows:
moles of aluminum = 1.00 g / 26.98 g/mol = 0.03704 mol
moles of hydrogen = 0.03704 mol x (3 mol H2 / 2 mol Al) = 0.05556 mol
mass of hydrogen = 0.05556 mol x 2.02 g/mol = 0.11212 g
So, approximately 0.11212 grams of hydrogen gas would be released by the reaction.
2KOH + Al + 6H2O--> 2KAl(OH)4 + 3H2
mols Al = grams/molar mass = ?
Use the coefficients in the balanced equation to convert mols Al to mols H2
Now convert mols H2 to grams. g = mols x molar mass