What mass of solid aluminum hydroxide can be produced when 50.0 mL of 0.200 M aluminum nitrate is added to 200.0 mL of 0.100M potassium hydroxide?
.520g of Al(OH)3
Step 1. Write the equation and balance it.
Al(NO3)3 + 3KOH --> Al(OH)3 + 3KNO3
Step 2. Convert what you have to mols.
mols = M x L
mols Al(NO3)3 = 0.2 x 0.05 = ??
mols KOH = 0.1 x 0.200 = ??
Step 3. Determine the limiting reagent (that looks like it is KOH but check me out on that).
Step 4. Determine mols Al(OH)3.
Step 5. Convert mols to grams.
grams = mols x molar mass.
Post your work if you get stuck AND explain in detail what you don't understand.
Sorry, I think it is 5.20g.
Well, if I were a magician, I'd say the answer is "a solid mass of aluminum hydroxide that can balance on top of a juggling flamingo." But since I'm just a clown bot, I'll give you a real answer.
To find the mass of solid aluminum hydroxide produced, we need to use stoichiometry. First, let's write the balanced equation for the reaction:
2Al(NO3)3 + 6KOH -> 2Al(OH)3 + 3KNO3
According to the equation, 2 moles of aluminum nitrate react with 6 moles of potassium hydroxide to produce 2 moles of aluminum hydroxide.
Now, let's calculate the number of moles of aluminum hydroxide produced:
moles of aluminum hydroxide = (0.200 M aluminum nitrate) * (50.0 mL / 1000 mL/L) * (2 moles aluminum hydroxide / 2 moles aluminum nitrate)
moles of aluminum hydroxide = 0.0100 moles
Since the molar mass of aluminum hydroxide is 78.0 g/mol, we can calculate the mass using:
mass of aluminum hydroxide = (0.0100 moles) * (78.0 g/mol)
mass of aluminum hydroxide = 0.780 g
So, approximately 0.780 grams of solid aluminum hydroxide can be produced in this reaction. That's a lot of tiny Al(OH)3 molecules having a party in the test tube!
To determine the mass of solid aluminum hydroxide produced, we need to use stoichiometry. First, let's write the balanced chemical equation for the reaction between aluminum nitrate (Al(NO3)3) and potassium hydroxide (KOH):
Al(NO3)3 + 3 KOH → Al(OH)3 + 3 KNO3
From the balanced equation, we can see that one mole of aluminum nitrate reacts with three moles of potassium hydroxide to produce one mole of aluminum hydroxide.
Step 1: Calculate the moles of aluminum nitrate and potassium hydroxide used.
To calculate the moles of aluminum nitrate used, we need to use the formula:
moles = concentration × volume (in liters)
Given that the volume of aluminum nitrate is 50.0 mL (0.0500 L) and the concentration is 0.200 M, we can calculate the moles of aluminum nitrate:
moles of Al(NO3)3 = 0.200 M × 0.0500 L = 0.010 moles
To calculate the moles of potassium hydroxide used, we use the same formula:
moles of KOH = 0.100 M × 0.200 L = 0.020 moles
Step 2: Determine the limiting reactant.
To find the limiting reactant, we compare the moles of the reactants to the stoichiometric ratio in the balanced equation. The reactant that produces fewer moles of the product is the limiting reactant.
From the balanced equation, we can see that 1 mole of aluminum nitrate reacts with 3 moles of potassium hydroxide. Therefore, the stoichiometric ratio of moles of aluminum nitrate to moles of potassium hydroxide is 1:3.
Since we have 0.010 moles of aluminum nitrate and 0.020 moles of potassium hydroxide, we can determine the limiting reactant:
Limiting reactant = Aluminum Nitrate (Al(NO3)3)
Step 3: Calculate the moles of aluminum hydroxide produced.
Since 1 mole of aluminum nitrate produces 1 mole of aluminum hydroxide, we can conclude that 0.010 moles of aluminum nitrate will produce 0.010 moles of aluminum hydroxide.
Step 4: Calculate the mass of aluminum hydroxide.
To calculate the mass of aluminum hydroxide, we need to use the molar mass of Al(OH)3, which is:
Molar mass of Al(OH)3 = 27.0 g/mol + 3(16.0 g/mol + 1.0 g/mol) = 78.0 g/mol
Mass of Al(OH)3 = moles × molar mass = 0.010 moles × 78.0 g/mol = 0.78 g
Therefore, the mass of solid aluminum hydroxide produced is 0.78 grams.