How many mL of 18 M H2SO4 are required to react with 250 mL of 2.50 M Al(OH)3 if the products are aluminum sulfate and water? How do i solve this?
250mL x 1L/1000mL x 2.50mol Al(OH)3/1L x 3 mol H2SO4/2 mol Al(OH)3 x 1L/18.0 mol H2SO4 x 1000mL/1L = 52.1mL H2SO4
1.Write the equation and balance it.
2.Convert 250 mL of 2.50 Al(OH)3 to moles. moles = M x L.
3. Using the coefficients in the balanced equation, convert moles Al(OH)3 to moles H2SO4.
4. Convert moles H2SO4 to volume. moles = M x L. Solve for L and convert to mL.
i. calculate the pH of the .833 sodium acetate solution
ii. Calculate the pH after adding .700 grams of NaOH to the 35 ml of .833 M sodium acetate solution
iii. Calculate the pH after adding 1.5 mL of 12 M HCl to the original 35 mL of .833 M sodium acetate solution
Well, let's dive into the chemistry clown car, shall we? To solve this, we need to use the balanced equation for the reaction between sulfuric acid (H2SO4) and aluminum hydroxide (Al(OH)3).
The balanced equation is:
2 Al(OH)3 + 3 H2SO4 → Al2(SO4)3 + 6 H2O
Now, to figure out how many mL of 18 M H2SO4 you need, we first need to determine the moles of Al(OH)3 present in the 250 mL of 2.50 M solution.
Moles = volume (in liters) × concentration (in moles/L)
So, moles of Al(OH)3 = 0.250 L × 2.50 mol/L
Next, let's use the stoichiometry from the balanced equation to find the moles of H2SO4 needed to react with the Al(OH)3.
From the balanced equation, we know that it takes 3 moles of H2SO4 to react with 2 moles of Al(OH)3.
Since 3:2 is the ratio, we can set up a proportion:
(moles of H2SO4 / moles of Al(OH)3) = (x / 2)
Now, solve for x (moles of H2SO4) to find out how many moles are required.
Finally, convert moles of H2SO4 to mL using the concentration given (18 M) and the volume (in L) you want to find.
mL of H2SO4 = (moles of H2SO4) × (1 L/18 mol)
Remember, this is just a clownish guide, and you should always double-check your calculations before proceeding with any reaction. Good luck, chemistry juggler!
To solve this question, you need to use the balanced chemical equation for the reaction between sulfuric acid (H2SO4) and aluminum hydroxide (Al(OH)3) to determine the stoichiometry of the reaction.
The balanced chemical equation for the reaction is:
H2SO4 + 2Al(OH)3 -> Al2(SO4)3 + 6H2O
The coefficients in the balanced equation represent the ratio of moles of each reactant and product. From the balanced equation, we can see that 1 mole of H2SO4 reacts with 2 moles of Al(OH)3 to produce 1 mole of Al2(SO4)3 and 6 moles of water.
To calculate the amount of H2SO4 needed to react with the given amount of Al(OH)3, we will use the concept of stoichiometry.
Step 1: Convert the given volume of Al(OH)3 to moles.
Using the given concentration, we can calculate the moles of Al(OH)3:
Moles of Al(OH)3 = 2.50 M x 0.250 L = 0.625 moles of Al(OH)3
Step 2: Determine the mole ratio between H2SO4 and Al(OH)3.
From the balanced equation, the mole ratio between H2SO4 and Al(OH)3 is 1:2.
Step 3: Calculate the moles of H2SO4.
Since the mole ratio between H2SO4 and Al(OH)3 is 1:2, the moles of H2SO4 is half of the moles of Al(OH)3:
Moles of H2SO4 = 0.625 moles of Al(OH)3 / 2 = 0.3125 moles of H2SO4
Step 4: Calculate the volume of 18 M H2SO4 solution.
Now, we need to convert the moles of H2SO4 to volume using the given concentration:
Volume of H2SO4 = Moles of H2SO4 / Concentration of H2SO4
Volume of H2SO4 = 0.3125 moles of H2SO4 / 18 M = 0.0174 L or 17.4 mL
Therefore, approximately 17.4 mL of 18 M H2SO4 are required to react with 250 mL of 2.50 M Al(OH)3.