What volume of 0.250M KOH is required to titrate 2.30 x 10^3 mol of the weak acid H2C2O4 ?

I know the answer is 18.4, but can you show the process?

The answer is 18.4 WHAT?

H2C2O4 + 2KOH ==> K2C2O4 + 2H2O
mols H2C2O4 = 2.30E3
mols KOH needed = twice that. Look at the coefficients. 1 mol H2C2O4 = 2 mols KOH.
M KOH = moles KOH/L KOH. Therefore,
L KOH = moles KOH/M KOH. I get 18,400 L. I wonder if you made a typo in the problem with mols 2.30E-3. If so the answer would be 0.0184L or 18.4 mL.

THX!

To find the volume of 0.250 M KOH required to titrate 2.30 x 10^3 mol of the weak acid H2C2O4, you can use the concept of stoichiometry and the balanced chemical equation for the reaction between KOH and H2C2O4.

The balanced chemical equation for the reaction is as follows:

2 KOH + H2C2O4 -> K2C2O4 + 2 H2O

From the balanced equation, you can see that 2 moles of KOH react with 1 mole of H2C2O4.

Given that you have 2.30 x 10^3 mol of H2C2O4 and the stoichiometric ratio is 2 moles of KOH to 1 mole of H2C2O4, you can calculate the moles of KOH required as follows:

(2.30 x 10^3 mol H2C2O4) x (2 mol KOH / 1 mol H2C2O4) = 4.60 x 10^3 mol KOH

Now, you need to use the molarity (concentration) of the KOH solution to find the volume of KOH required. The formula to calculate volume is:

Volume (in liters) = moles / molarity

Substituting the values, you get:

(4.60 x 10^3 mol KOH) / (0.250 mol/L) = 18.4 L

Therefore, the volume of 0.250 M KOH required to titrate 2.30 x 10^3 mol of H2C2O4 is 18.4 liters.

To find the volume of 0.250M KOH required to titrate 2.30 x 10^3 mol of the weak acid H2C2O4, we can use the concept of stoichiometry.

First, we need to determine the balanced chemical equation for the reaction between KOH and H2C2O4. The balanced equation is:

2 KOH + H2C2O4 -> K2C2O4 + 2 H2O

From the equation, we can see that 2 moles of KOH react with 1 mole of H2C2O4. This means that the ratio of KOH to H2C2O4 is 2:1.

Next, we can use the balanced equation to calculate the moles of KOH needed to react with the given moles of H2C2O4.

Moles of H2C2O4 = 2.30 x 10^3 mol

We know that the ratio of KOH to H2C2O4 is 2:1, so the moles of KOH required can be calculated as:

Moles of KOH = (Moles of H2C2O4) / 2
= (2.30 x 10^3 mol) / 2
= 1.15 x 10^3 mol

Now, we need to convert the moles of KOH to volume using the given concentration of KOH.

Concentration of KOH = 0.250 M

We can use the formula:

Volume (in liters) = Moles / Concentration

Volume of KOH = (Moles of KOH) / (Concentration of KOH)
= (1.15 x 10^3 mol) / (0.250 M)
= 4600 L

However, we are asked to provide the volume in cubic decimeters (dm^3). So we need to convert the volume from liters to cubic decimeters.

1 L = 1 dm^3, so

Volume of KOH = 4600 L = 4600 dm^3

Finally, we can round the volume to the appropriate number of significant figures.

Volume of KOH = 18.4 dm^3

Therefore, the volume of 0.250 M KOH required to titrate 2.30 x 10^3 mol of H2C2O4 is 18.4 dm^3.