Determine the volume of a 0.550 M KMnO4 solution required to completely react with 2.80 g of Zn.

1. Write the equation and balance it.

2. Calculate moles Zn. mol =- g/molar mass
3. Using the coefficients in the balanced equation, convert mols Zn to mols KMnO4.
4. The M KMnO4 = moles KMnO4/L KMnO4.
You knolw M and moles, solve for L and convert to mL.

To determine the volume of the KMnO4 solution required to completely react with 2.80 g of Zn, we will use the balanced chemical equation for the reaction between KMnO4 and Zn:

2 KMnO4 + 3 Zn -> 2 MnO2 + 2 KOH + ZnO

From the balanced equation, we can see that each mole of KMnO4 reacts with 3 moles of Zn.

Step 1: Calculate the number of moles of Zn
To calculate the number of moles of Zn, we will use its molar mass.

molar mass of Zn = 65.38 g/mol

moles of Zn = mass of Zn / molar mass of Zn = 2.80 g / 65.38 g/mol

Step 2: Determine the volume of KMnO4 solution
To determine the volume of KMnO4 solution required, we need to calculate the number of moles of KMnO4 using the mole ratio from the balanced equation.

From the balanced equation, we know that 2 moles of KMnO4 react with 3 moles of Zn.

moles of KMnO4 = (moles of Zn) x (2 moles of KMnO4 / 3 moles of Zn)

Step 3: Calculate the volume of KMnO4 solution
To calculate the volume of KMnO4 solution, we will use the molarity of the solution.

Molarity (M) = moles / volume (L)

We need to rearrange the equation to solve for volume:

volume (L) = moles / Molarity (M)

Given that the molarity is 0.550 M, we can substitute the values and calculate the volume:

volume = (moles of KMnO4) / (Molarity of KMnO4 solution)

Now let's substitute the values and calculate the volume:

volume = (moles of Zn) x (2 moles of KMnO4 / 3 moles of Zn) / (0.550 mol/L)

After calculating this expression, you will obtain the volume of the KMnO4 solution required to completely react with 2.80 g of Zn.

To determine the volume of the KMnO4 solution required to react with 2.80 g of Zn, we need to use the stoichiometry of the balanced chemical equation between KMnO4 and Zn.

First, let's write the balanced chemical equation for the reaction:
2 KMnO4 + 3 Zn → 2 MnO2 + 2 KOH + ZnO

From the balanced equation, we can see that 2 moles of KMnO4 react with 3 moles of Zn.

To find the moles of Zn, we can use the molar mass of Zn:
Molar mass of Zn = 65.38 g/mol
2.80 g Zn × (1 mol Zn / 65.38 g Zn) = 0.0428 mol Zn

Using the stoichiometry, we know that 2 moles of KMnO4 react with 3 moles of Zn.

So, if 3 moles of Zn react, we can calculate the moles of KMnO4 required:
0.0428 mol Zn × (2 mol KMnO4 / 3 mol Zn) = 0.0285 mol KMnO4

Finally, we can calculate the volume of the KMnO4 solution using the molarity (0.550 M = 0.550 mol/L) and the calculated moles:
Volume = Moles / Molarity
Volume = 0.0285 mol KMnO4 / 0.550 mol/L = 0.0518 L

The volume of the 0.550 M KMnO4 solution required to completely react with 2.80 g of Zn is 0.0518 liters (or 51.8 mL).

32.8mL