A sample of g 0.2050 oxalic acid (4.2 H2C2O H2O) required 25.52 mL of KOH solution to complete the neutralization according to rea

tion: HOOCCOOH + 2 KOH + KOOCCOOK ↔ H2O Which the concentration in mol L-1 of this KOH solution? A sample of an unknown oxalate 0.3025 g required 22.50 mL the same solution of KOH, what percentage expressed in H2C2O4 in the sample?

What is that 4.2 H2C2O4?).

To find the concentration of the KOH solution in mol/L, we can use the balanced equation and stoichiometry. The balanced equation is as follows:

HOOCCOOH + 2KOH + KOOCCOOK ↔ H2O

From the equation, we can see that 1 mole of oxalic acid (HOOCCOOH) reacts with 2 moles of KOH. Therefore, the number of moles of oxalic acid used in the reaction can be calculated as follows:

moles of oxalic acid = mass of oxalic acid (g) / molar mass of oxalic acid (g/mol)

Given that the mass of the oxalic acid is 0.2050 g and the molar mass of oxalic acid is 90.03 g/mol (4.2 H2C2O H2O), we can calculate the moles of oxalic acid:

moles of oxalic acid = 0.2050 g / 90.03 g/mol = 0.00227 mol

Since 2 moles of KOH react with 1 mole of oxalic acid, the moles of KOH used in the reaction can be calculated as follows:

moles of KOH = 0.00227 mol * 2 = 0.00454 mol

Now, we can calculate the concentration of the KOH solution in mol/L using the formula:

concentration (mol/L) = moles of solute / volume of solution (L)

Given that the volume of the KOH solution is 25.52 mL (which is 0.02552 L), we can substitute the values into the formula:

concentration (mol/L) = 0.00454 mol / 0.02552 L ≈ 0.178 mol/L

Therefore, the concentration of the KOH solution is approximately 0.178 mol/L.

Moving on to the second part of the question:
To find the percentage of oxalic acid in the unknown oxalate sample, we need to determine the moles of oxalic acid in the sample and compare it to the mass of the sample.

First, calculate the moles of KOH that reacted with the unknown oxalate sample using the same process as before. The moles of KOH can be calculated as follows:

moles of KOH = concentration (mol/L) * volume of KOH (L)
= 0.178 mol/L * 0.02250 L
= 0.004 mmol

Next, since 1 mole of oxalic acid reacts with 2 moles of KOH (as per the balanced equation), we can calculate the moles of oxalic acid in the sample:

moles of oxalic acid = 0.004 mmol * (1 mol oxalic acid / 2 mol KOH)
= 0.002 mmol

Given that the mass of the unknown oxalate sample is 0.3025 g, we can calculate the percentage of oxalic acid in the sample:

% oxalic acid = (moles of oxalic acid * molar mass of oxalic acid) / mass of sample * 100

Substituting the values into the formula:

% oxalic acid = (0.002 mmol * 90.03 g/mol) / 0.3025 g * 100
= 5.94%

Therefore, the percentage of oxalic acid in the unknown oxalate sample is approximately 5.94%.