Question

What mass of \(\mathrm{KMnO}_{4}\) is needed to react with \(100, \mathrm{~mL}, \mathrm{H}_{2} \mathrm{O}_{2}\) solution? ( \(a=1,031 \mathrm{~g} / \mathrm{mL}, 9.0^{3} \% \mathrm{H}_{2} \mathrm{O}_{2}\) by mass) $$ \begin{aligned} \mathrm{H}_{2} \mathrm{O}_{2}+& \mathrm{KMnO}_{4}+\mathrm{H}_{2} \mathrm{SO}_{4} \longrightarrow \\ & \mathrm{O}_{2}+\mathrm{MnSO}_{4}+\mathrm{K}_{2} \mathrm{SO}_{4}+\mathrm{H}_{2} \mathrm{O} \quad \text { (acídic solution) } \end{aligned} $$

Short answer

17.23 g of \text{KMnO}_{4}

Step-by-step solution

Determine the mass of \text{H}_{2}\text{O}_{2} in the solution

First, determine the mass of \text{H}_{2}\text{O}_{2} in the solution. The volume of the solution given is 100 mL. The density (\text{a}) provided is 1.031 g/mL and the percentage mass of \text{H}_{2}\text{O}_{2} is 9.0%. This means the total mass of the solution is: \[ \text{mass}_{\text{solution}} = \text{volume} \times \text{density} = 100 \text{ mL} \times 1.031 \text{ g/mL} = 103.1 \text{ g} \] Then, the mass of \text{H}_{2}\text{O}_{2} is: \[ \text{mass}_{\text{H}_{2}\text{O}_{2}} = 0.09 \times 103.1 \text{ g} = 9.279 \text{ g} \]

Convert mass of \text{H}_{2}\text{O}_{2} to moles

Next, use the molar mass of \text{H}_{2}\text{O}_{2} to convert the mass to moles. The molar mass of \text{H}_{2}\text{O}_{2} is 34.01 g/mol. The number of moles of \text{H}_{2}\text{O}_{2} is: \[ \text{moles of } \text{H}_{2}\text{O}_{2} = \frac{9.279 \text{ g}}{34.01 \text{ g/mol}} = 0.273 \text{ moles} \]

Write and balance the chemical equation

The balanced chemical equation for the reaction is: \[ 5 \text{H}_{2}\text{O}_{2} + 2 \text{KMnO}_{4} + 3 \text{H}_{2}\text{SO}_{4} \rightarrow 5 \text{O}_{2} + 2 \text{MnSO}_{4} + \text{K}_{2}\text{SO}_{4} + 8 \text{H}_{2} \text{O} \] According to the equation, 5 moles of \text{H}_{2}\text{O}_{2} react with 2 moles of \text{KMnO}_{4}.

Calculate the moles of \text{KMnO}_{4} needed

Using the stoichiometric coefficients from the balanced equation, determine the moles of \text{KMnO}_{4} that react with 0.273 moles of \text{H}_{2}\text{O}_{2}. The mole ratio from \text{H}_{2}\text{O}_{2} to \text{KMnO}_{4} is 5:2. So: \[ \text{moles of } \text{KMnO}_{4} = \frac{2}{5} \times 0.273 \text{ moles} = 0.109 \text{ moles} \]

Convert moles of \text{KMnO}_{4} to grams

Finally, convert the moles of \text{KMnO}_{4} needed to grams. The molar mass of \text{KMnO}_{4} is 158.04 g/mol. Thus: \[ \text{mass of } \text{KMnO}_{4} = 0.109 \text{ moles} \times 158.04 \text{ g/mol} = 17.23 \text{ g} \]

Key concepts

chemical reactions

A chemical reaction is a process where reactants are transformed into products. In the given exercise, the reaction involves several compounds: \text{H}_{2}\text{O}_{2}, \text{KMnO}_{4}, and \text{H}_{2}\text{SO}_{4}. When they react, new substances like \text{O}_{2}, \text{MnSO}_{4}, \text{K}_{2}\text{SO}_{4}, and \text{H}_{2}\text{O} are formed.
This process involves breaking old bonds and forming new ones, and is represented by a chemical equation. The equation must be balanced, meaning the number of atoms of each element on both sides must be equal. This ensures the law of conservation of mass is obeyed.
For the given reaction, the balanced equation is:
\[5 \text{H}_{2}\text{O}_{2} + 2 \text{KMnO}_{4} + 3 \text{H}_{2}\text{SO}_{4} \rightarrow 5 \text{O}_{2} + 2 \text{MnSO}_{4} + \text{K}_{2}\text{SO}_{4} + 8 \text{H}_{2} \text{O} \]

molar mass

Molar mass is the mass of one mole of a substance, typically expressed in grams per mole (g/mol). It's a crucial concept for converting between mass and moles.
To find the molar mass, sum the atomic masses of all atoms in the molecule. For \text{H}_{2}\text{O}_{2}, the calculation would be:
\[2 \times 1.01 \text{ (H)} + 2 \times 16.00 \text{ (O)} = 34.01 \text{ g/mol}\]Similarly, for \text{KMnO}_{4}, the molar mass is calculated as:
\[39.10 \text{ (K)} + 54.94 \text{ (Mn)} + 4 \times 16.00 \text{ (O)} = 158.04 \text{ g/mol}\]These calculations allow us to convert grams of a substance to moles and vice versa, which is essential in stoichiometric calculations.

stoichiometric calculations

Stoichiometry involves calculating the amounts of reactants and products in a chemical reaction. It uses the mole ratio derived from the balanced chemical equation.
For example, the balanced equation in our problem indicates that 5 moles of \text{H}_{2}\text{O}_{2} react with 2 moles of \text{KMnO}_{4}. This ratio (5:2) allows us to determine how much \text{KMnO}_{4} is required to react completely with a given amount of \text{H}_{2}\text{O}_{2}.
In our calculation, we found 0.273 moles of \text{H}_{2}\text{O}_{2}. Using the mole ratio:
\[ \text{moles of } \text{KMnO}_{4} = \frac{2}{5} \times 0.273 = 0.109 \text{ moles}\] Finally, we convert moles of \text{KMnO}_{4} to grams:
\[0.109 \text{ moles} \times 158.04 \text{ g/mol} = 17.23 \text{ g}\]

balancing chemical equations

Balancing chemical equations is essential to ensure that the number of atoms of each element is the same on both sides of the reaction. This reflects the conservation of mass.
Start by writing the unbalanced equation and count the number of each type of atom on both sides. Adjust coefficients (the numbers before molecules) to balance the atoms.
For the given reaction:
\text{Unbalanced:}
\[ \text{H}_{2}\text{O}_{2} + \text{KMnO}_{4} + \text{H}_{2}\text{SO}_{4} \rightarrow \text{O}_{2} + \text{MnSO}_{4} + \text{K}_{2}\text{SO}_{4} + \text{H}_{2}\text{O} \]
Balanced:
\[5 \text{H}_{2}\text{O}_{2} + 2 \text{KMnO}_{4} + 3 \text{H}_{2}\text{SO}_{4} \rightarrow 5 \text{O}_{2} + 2 \text{MnSO}_{4} + \text{K}_{2}\text{SO}_{4} + 8 \text{H}_{2}\text{O}\]
Notice how each type of atom balances across the reaction side and product side. This method ensures that the reaction adheres to the law of conservation of mass.