how many millimeters of 0.200M FeCL3 are needed to react with an excess of Na2S to produce 1.38g of Fe2S3 if the percent yield for the reaction is 65.0%

Well, let me calculate that for you! But before I do, just remember that I'm not a math expert, so take my answer with a grain of salt - or maybe a whole shaker of salt, depending on your taste.

To start, we need to find the molar mass of Fe2S3. You could think of it as two Fe atoms and three S atoms going on a fancy date, but chemistry-wise, it's 2*55.85 g/mol (Fe) + 3*32.06 g/mol (S) = 207.72 g/mol.

Now, we can find the number of moles of Fe2S3 produced using the given mass: 1.38 g / 207.72 g/mol = 0.00664 mol (approximately).

Since the reaction has a 65.0% yield, we can find the theoretical amount of FeCl3 needed. Theoretical yield = 0.00664 mol divided by 0.65 = 0.0102 mol (approximately).

To find the volume of 0.200 M FeCl3 needed, we divide the moles by the molarity: 0.0102 mol / 0.200 mol/L = 0.051 L or 51.0 mL (since 1 L = 1000 mL).

So, approximately 51.0 milliliters of 0.200M FeCl3 are needed. But remember, this answer isn't perfect - it's just like my attempts at stand-up comedy.

To determine the amount of 0.200M FeCl3 needed, we need to calculate the stoichiometric ratio between FeCl3 and Fe2S3.

The balanced equation for the reaction between FeCl3 and Na2S is as follows:

2FeCl3 + 3Na2S -> Fe2S3 + 6NaCl

From the equation, we can see that 2 moles of FeCl3 react to produce 1 mole of Fe2S3.

Step 1: Calculate the moles of Fe2S3 produced

Given mass of Fe2S3 = 1.38g
Molar mass of Fe2S3 = 55.85g/mol + 32.07g/mol = 87.92g/mol

moles of Fe2S3 = mass / molar mass
= 1.38g / 87.92g/mol
≈ 0.0157 mol

Step 2: Calculate the moles of FeCl3 needed

From the stoichiometric ratio, we know that 2 moles of FeCl3 are required to produce 1 mole of Fe2S3.

moles of FeCl3 needed = 2 * moles of Fe2S3
= 2 * 0.0157 mol
≈ 0.0314 mol

Step 3: Calculate the volume of 0.200M FeCl3 needed

Molarity (M) = moles / volume
0.200M = 0.0314 mol / volume

volume = moles / Molarity
= 0.0314 mol / 0.200 mol/L
≈ 0.157 L

Step 4: Convert the volume to milliliters

1 L = 1000 mL

volume (mL) = 0.157 L * 1000 mL/L
= 157 mL

Therefore, approximately 157 milliliters of 0.200M FeCl3 are needed to react with an excess of Na2S to produce 1.38g of Fe2S3, assuming a percent yield of 65.0%.

To determine the number of millimeters of 0.200M FeCl3 needed, we first need to calculate the amount of FeCl3 required to react completely with Na2S and then take into account the percent yield.

Step 1: Write and balance the chemical equation for the reaction:
2FeCl3 + 3Na2S → Fe2S3 + 6NaCl

Step 2: Calculate the molar mass of Fe2S3:
Fe2S3 = (55.85 x 2) + (32.06 x 3) = 207.4 g/mol

Step 3: Convert the given mass of Fe2S3 to moles:
1.38 g Fe2S3 x (1 mol Fe2S3 / 207.4 g Fe2S3) = 0.00665 mol Fe2S3

Step 4: Since the balanced equation has a 1:1 stoichiometric ratio between Fe2S3 and FeCl3, the moles of FeCl3 required will be the same as the moles of Fe2S3.

Step 5: Calculate the theoretical yield of FeCl3:
Theoretical yield = 0.00665 mol FeCl3

Step 6: Take into account the percent yield:
Percent yield = (Actual yield / Theoretical yield) x 100

Rearrange the formula to solve for actual yield:
Actual yield = (Percent yield / 100) x Theoretical yield

Actual yield = (65.0 / 100) x 0.00665 mol FeCl3 ≈ 0.00433 mol FeCl3

Step 7: Calculate the volume using the molarity of FeCl3 solution:
Molarity (M) = moles / volume (L)
0.200 M = 0.00433 mol / volume (L)

Rearrange the formula to solve for volume:
Volume (L) = moles / Molarity

Volume (L) = 0.00433 mol / 0.200 M ≈ 0.0217 L

Step 8: Convert volume to milliliters:
1 L = 1000 mL

Volume (mL) = 0.0217 L x 1000 mL/L ≈ 21.7 mL

Therefore, approximately 21.7 milliliters of 0.200M FeCl3 are needed to react with an excess of Na2S and produce 1.38g of Fe2S3 if the percent yield for the reaction is 65.0%.

2FeCl3 + 3Na2S ==> Fe2S3 + 6NaCl

mols Fe2S3 = grams/molar mass = ? That's the mols you want if the reaction were 100% but it is only 65%; therefore, you want ? mol/0.65 = mols needed at 65% yield.
Using the coefficients in the balanced equation, convert mols Fe2S3 to mols
FeCl3.
Then M FeCl3 = mols FeCl3/L FeCl3. You know M and mols, solve for L and convert to mL.