Determine the volume of a 0.550 M KMnO4 solution required to completely react with 2.55 g of Zn.

Oh, I see you're trying to make me calculate. Well, I must say I'm not a big fan of math, but I'll give it a shot for you. Just please don't expect any funny jokes about numbers!

To determine the volume of the KMnO4 solution required to react with 2.55 g of Zn, we need to use stoichiometry.

First, let's convert the mass of Zn to moles. The molar mass of zinc (Zn) is approximately 65.38 g/mol. So, 2.55 g of Zn is equal to 2.55 g × (1 mol/65.38 g) ≈ 0.039 mol.

According to the balanced equation, 1 mol of KMnO4 reacts with 1 mol of Zn. So, the stoichiometric ratio is 1:1.

Now, since we know the concentration of the KMnO4 solution is 0.550 M, we can use the formula:

moles = concentration × volume

Rearranging the equation to solve for volume, we get:

volume = moles / concentration

Plugging in our values:

volume = 0.039 mol / 0.550 M ≈ 0.071 L

So, the volume of the 0.550 M KMnO4 solution required to completely react with 2.55 g of Zn is approximately 0.071 liters.

To determine the volume of the KMnO4 solution required to react with the given amount of Zn, we need to apply stoichiometry. The balanced equation for the reaction between KMnO4 and Zn is:

2 KMnO4 + 8H2SO4 + 10Zn → 5ZnSO4 + 2MnSO4 + K2SO4 + 8H2O

From the equation, we can see that 2 moles of KMnO4 react with 10 moles of Zn. Therefore, we need to calculate the number of moles of Zn present in 2.55 g.

Step 1: Calculate the number of moles of Zn:
Molar mass of Zn = 65.38 g/mol
Moles of Zn = mass / molar mass = 2.55 g / 65.38 g/mol = 0.039 moles

Step 2: Determine the number of moles of KMnO4 required:
From the balanced equation, the stoichiometric ratio of KMnO4 to Zn is 2:10.
So, the moles of KMnO4 required = (0.039 moles Zn) x (2 moles KMnO4 / 10 moles Zn) = 0.0078 moles KMnO4.

Step 3: Convert moles of KMnO4 to volume (in liters):
Molarity (M) = Moles / Volume (L)
0.550 M = 0.0078 moles KMnO4 / Volume (L)

Rearranging the equation:
Volume (L) = 0.0078 moles KMnO4 / 0.550 M = 0.0142 L = 14.2 mL

Therefore, the volume of the 0.550 M KMnO4 solution required to completely react with 2.55 g of Zn is 14.2 mL.

To determine the volume of the KMnO4 solution required to completely react with the given amount of Zn, we need to set up and solve a stoichiometry problem using the balanced chemical equation representing the reaction between KMnO4 and Zn.

The balanced chemical equation for the reaction is:
2 KMnO4 + 10 Zn → 2 MnO2 + 5 ZnO + K2ZnO2

From the balanced equation, we can see that 2 moles of KMnO4 react with 10 moles of Zn.

Step 1: Convert the mass of Zn to moles.

The molar mass of Zn is 65.38 g/mol.
Using the given mass of Zn (2.55 g), we can calculate the number of moles:

moles of Zn = mass of Zn / molar mass of Zn
moles of Zn = 2.55 g / 65.38 g/mol = 0.039 moles of Zn

Step 2: Determine the volume of KMnO4 solution.

Since the stoichiometric ratio between KMnO4 and Zn is 2:10, we can set up a proportion:

moles of KMnO4 / volume of KMnO4 = moles of Zn / stoichiometric coefficient of Zn

From the proportion, we can determine the volume of KMnO4 solution:

volume of KMnO4 = (moles of KMnO4 * volume of KMnO4) / moles of Zn

However, we still need to find the moles of KMnO4.

Step 3: Use the molarity (0.550 M) and the volume (the unknown) to find moles of KMnO4.

Molarity (M) is defined as moles of solute divided by liters of solution. Rearranging the formula, we have:

moles of solute = Molarity * volume of solution

From the molarity and volume of KMnO4, we can calculate the moles of KMnO4:

moles of KMnO4 = Molarity * volume of KMnO4

Step 4: Substitute the moles of KMnO4 into the volume expression.

Replace the calculated value of moles of KMnO4 into the volume expression from Step 2:

volume of KMnO4 = (moles of KMnO4 * volume of KMnO4) / moles of Zn

Now you can substitute the values into the equation and solve for the volume of KMnO4:

volume of KMnO4 = (0.550 M * volume of KMnO4) / 0.039 moles

Since the volume of KMnO4 is the unknown, we need to isolate it on one side of the equation. Multiply both sides of the equation by 0.039 moles:

volume of KMnO4 * 0.039 moles = 0.550 M * volume of KMnO4

Now divide both sides of the equation by 0.550 M:

volume of KMnO4 * 0.039 moles / 0.550 M = volume of KMnO4

Evaluate the expression:

volume of KMnO4 = (0.039 moles * 1 L * 0.550 M) / (0.550 M)

volume of KMnO4 = 1 L

Therefore, the volume of the 0.550 M KMnO4 solution required to completely react with 2.55 g of Zn is 1 liter.

Write and balance the equation.

Convert 2.55 g Zn to moles.
Using the coefficients in the balanced equation, convert moles Zn to moles KMnO4.
Then M = moles/L. You know M and moles, solve for L.