Determine the volume of a 0.550M of kmno4 solution required to completely react with 2.70 of zn

2.70 WHAT of Zn. grams?

Write and balance the equation.
moles Zn = grams/molar mass
Using the coefficients in the balanced equation, convert moles Zn to moles KMnO4.
Then M KMnO4 = moles/L
You know M and moles, solve for L

To determine the volume of the KMnO4 solution required to completely react with 2.70g of Zn, we need to use the stoichiometry of the balanced chemical equation.

The balanced equation for the reaction between KMnO4 and Zn is:

2 KMnO4 + 3 Zn → 2 MnO2 + 3 ZnO + 2 K2O

From the balanced equation, we can see that the mole ratio between KMnO4 and Zn is 2:3. This means that for every 2 moles of KMnO4, we need 3 moles of Zn.

First, we need to convert the mass of Zn to moles. The molar mass of Zn is 65.38 g/mol, so:

2.70 g Zn * (1 mol Zn / 65.38 g Zn) = 0.04132 mol Zn

Next, we can determine the moles of KMnO4 needed using the mole ratio from the balanced equation:

0.04132 mol Zn * (2 mol KMnO4 / 3 mol Zn) = 0.0275 mol KMnO4

Now we have the number of moles of KMnO4 required. To find the volume of the 0.550M KMnO4 solution needed, we can use the equation:

Moles = Molarity * Volume

0.0275 mol KMnO4 = 0.550 M * Volume

Rearranging the equation, we find:

Volume = Moles / Molarity

Volume = 0.0275 mol / 0.550 M = 0.050 L

To convert from liters to milliliters, we multiply by 1000:

Volume = 0.050 L * 1000 mL/L = 50 mL

Therefore, the volume of the 0.550M KMnO4 solution required to completely react with 2.70g of Zn is 50 mL.