What volume of 0.200 M K2C2O4 is required to completely react with 30.0 mL of 0.100 M Fe(NO3)3?

2Fe(NO3)3 + 3K2C2O4 --> Fe2(C2O4)3 + 6KNO3

To answer this question, we need to determine the stoichiometry of the reaction. From the balanced equation, we can see that 2 moles of Fe(NO3)3 react with 3 moles of K2C2O4.

First, let's calculate the number of moles of Fe(NO3)3 in the given volume:

moles of Fe(NO3)3 = volume x molarity
moles of Fe(NO3)3 = 30.0 mL x 0.100 mol/L (since 0.100 M is equivalent to 0.100 mol/L)
moles of Fe(NO3)3 = 3.00 mmol (since 30.0 mL is equivalent to 3.00 mmol)

Now, let's use the stoichiometry of the reaction to calculate the amount of K2C2O4 needed. According to the stoichiometry, we need 3 moles of K2C2O4 for every 2 moles of Fe(NO3)3.

moles of K2C2O4 = (2/3) x moles of Fe(NO3)3
moles of K2C2O4 = (2/3) x 3.00 mmol
moles of K2C2O4 = 2.00 mmol

Next, we need to convert the amount of K2C2O4 from moles to volume. To do this, we'll use the molarity of K2C2O4:

volume of K2C2O4 = moles of K2C2O4 / molarity
volume of K2C2O4 = 2.00 mmol / 0.200 mol/L (since 0.200 M is equivalent to 0.200 mol/L)
volume of K2C2O4 = 10.0 mL (since 2.00 mmol is equivalent to 10.0 mL)

Therefore, 10.0 mL of the 0.200 M K2C2O4 solution is required to completely react with 30.0 mL of the 0.100 M Fe(NO3)3 solution.

To determine the volume of 0.200 M K2C2O4 required to completely react with 30.0 mL of 0.100 M Fe(NO3)3, we can use the concept of stoichiometry and the given balanced equation.

First, let's calculate the number of moles of Fe(NO3)3 in 30.0 mL of the solution using the formula:

moles = concentration × volume

moles of Fe(NO3)3 = 0.100 M × 0.030 L = 0.003 moles

According to the balanced equation, the stoichiometric ratio between Fe(NO3)3 and K2C2O4 is 2:3, which means that for every 2 moles of Fe(NO3)3, we will need 3 moles of K2C2O4.

So, using the stoichiometry, we can calculate the number of moles of K2C2O4 required:

moles of K2C2O4 = (3/2) × moles of Fe(NO3)3
= (3/2) × 0.003
= 0.0045 moles

Now, we need to determine the volume of 0.200 M K2C2O4 required to have 0.0045 moles. We can use the formula:

volume = moles / concentration

volume of K2C2O4 = 0.0045 moles / 0.200 M
= 0.0225 L
= 22.5 mL

Therefore, 22.5 mL of 0.200 M K2C2O4 is required to completely react with 30.0 mL of 0.100 M Fe(NO3)3.

mols Fe(NO3)3 = M x L = ?

Use the coefficients in the balanced equation to convert mols Fe(NO3)3 to mols K2C2O4.
Then M K2C2O4 = mols K2C2O4/L K2C2O4. You know mols and M, solve for L and convert to mL if needed. Post your work if you get stuck.