What volume of 0.0500 M Ba(OH)2 will react completely with 27.50 mL of 0.550 M HCl?

Ba(OH)2 + 2HCl ==> BaCl2 + 2H2O

mols HCl = M x L
Mols Ba(OH)2 = mols HCl x (1 mol Ba(OH)2)/2 mols HCl).
M Ba(OH)2 = mol/L. You have M and you have mols, calculate L. Check my work. It's getting late.

Well, let me do some calculations while trying to keep the mood light!

First, we need to balance the equation for the reaction between Ba(OH)2 and HCl: Ba(OH)2 + 2HCl -> BaCl2 + 2H2O.

Now, let's find the amount of moles in 27.50 mL of 0.550 M HCl. We can use the formula Molarity = Moles/Liter (M = mol/L).

So, moles of HCl = 0.550 mol/L * 0.02750 L = 0.0151 moles.

Since the stoichiometric ratio of Ba(OH)2 to HCl is 1:2, we need twice as many moles of Ba(OH)2 to react completely.

Therefore, the total moles of Ba(OH)2 needed = 2 * 0.0151 moles = 0.0302 moles.

Now, we can find the volume of 0.0500 M Ba(OH)2 that contains 0.0302 moles. Using the same formula, Volume = Moles/Molarity (V = mol/M), we have:

Volume = 0.0302 moles / 0.0500 mol/L = 0.604 L = 604 mL.

So, to answer your question, you would need approximately 604 mL of 0.0500 M Ba(OH)2 to completely react with 27.50 mL of 0.550 M HCl. But please note that my calculations are just for humor, not for official scientific accuracy!

To determine the volume of 0.0500 M Ba(OH)2 needed to react completely with 27.50 mL of 0.550 M HCl, we can use stoichiometry and the concept of molarity.

First, let's write the balanced chemical equation for the reaction between Ba(OH)2 and HCl:

Ba(OH)2 + 2HCl -> BaCl2 + 2H2O

From the balanced equation, we can see that 1 mole of Ba(OH)2 reacts with 2 moles of HCl.

Step 1: Convert the given volume of HCl from mL to L:
27.50 mL = 27.50/1000 = 0.02750 L

Step 2: Use the equation Molarity = Moles/Liters to calculate the number of moles of HCl:
Moles of HCl = Molarity × Volume
Moles of HCl = 0.550 M × 0.02750 L = 0.0151 moles

Step 3: Use stoichiometry to determine the number of moles of Ba(OH)2 needed to react with the HCl.
From the balanced equation, we can see that 1 mole of Ba(OH)2 reacts with 2 moles of HCl.
Therefore, the moles of Ba(OH)2 = 0.0151/2 = 0.00755 moles

Step 4: Use the equation Molarity = Moles/Liters to calculate the volume of 0.0500 M Ba(OH)2 needed to react with the calculated moles of Ba(OH)2:
Volume of Ba(OH)2 = Moles/Molarity
Volume of Ba(OH)2 = 0.00755 moles / 0.0500 M = 0.151 L

Step 5: Convert the volume of Ba(OH)2 from liters to milliliters:
Volume of Ba(OH)2 = 0.151 L × 1000 = 151 mL

Therefore, 151 mL of 0.0500 M Ba(OH)2 will react completely with 27.50 mL of 0.550 M HCl.

To find the volume of 0.0500 M Ba(OH)2 needed to react completely with 27.50 mL of 0.550 M HCl, we can use the concept of stoichiometry and the balanced chemical equation.

The balanced chemical equation for the reaction between Ba(OH)2 and HCl is:

Ba(OH)2 + 2HCl → BaCl2 + 2H2O

From the balanced equation, we can see that 1 mole of Ba(OH)2 reacts with 2 moles of HCl.

First, we need to find the number of moles of HCl in 27.50 mL of 0.550 M HCl:

Number of moles of HCl = Volume (in L) × Concentration (in M)
= 27.50 mL × 0.550 M / 1000 mL/L
= 0.015125 mol

According to the stoichiometry of the balanced equation, 1 mole of Ba(OH)2 reacts with 2 moles of HCl. Therefore, the number of moles of Ba(OH)2 needed can be calculated as follows:

Number of moles of Ba(OH)2 needed = (0.015125 mol HCl) / 2
= 0.0075625 mol

Finally, to find the volume of 0.0500 M Ba(OH)2 needed, we can use the molarity and the number of moles calculated:

Volume (in L) = Number of moles / Concentration
= 0.0075625 mol / 0.0500 M
= 0.15125 L

Since the volume was initially given in mL, we convert it back:

Volume (in mL) = 0.15125 L × 1000 mL/L
= 151.25 mL

Therefore, approximately 151.25 mL of 0.0500 M Ba(OH)2 is needed to react completely with 27.50 mL of 0.550 M HCl.