Question

What mass of \(3.00 \% \mathrm{H}_{2} \mathrm{O}_{2}\) solution is needed to produce \(35.7 \mathrm{~g}\) of \(\mathrm{O}_{2}(\mathrm{~g})\) at \(295 \mathrm{~K}\) at \(1.05 \mathrm{~atm}\) pressure? $$ 2 \mathrm{H}_{2} \mathrm{O}_{2}(\mathrm{aq}) \rightarrow 2 \mathrm{H}_{2} \mathrm{O}(\ell)+\mathrm{O}_{2}(\mathrm{~g}). $$

Short answer

Around 2530 grams of solution are needed.

Step-by-step solution

Determine Moles of Oxygen

First, we use the ideal gas law to find the moles of oxygen needed. The ideal gas law is given by \( PV = nRT \). Here, \( P = 1.05 \, \text{atm} \), \( V \) is the unknown volume, but we are given mass instead, \( R = 0.0821 \, \text{L} \cdot \text{atm/mol} \cdot \text{K} \), and \( T = 295 \, \text{K} \). We start by calculating the moles of \( \mathrm{O}_2 \). Rearranging the ideal gas law to solve for \( n \), we have: \[ n = \frac{PV}{RT} \implies n = \frac{1.05 \cdot V}{0.0821 \times 295} \]. However, we can use the mass of \( \mathrm{O}_2 \) directly to find the moles, as \( \text{mass} = n \times \text{molar mass} \), thus:\[ n = \frac{35.7}{32.00} = 1.116 \text{ moles of } \mathrm{O}_2 \].

Use Reaction Stoichiometry

According to the balanced equation \[ 2 \mathrm{H}_2\mathrm{O}_2 \rightarrow 2 \mathrm{H}_2\mathrm{O} + \mathrm{O}_2 \], every two moles of \( \mathrm{H}_2\mathrm{O}_2 \) produce one mole of \( \mathrm{O}_2 \). Therefore, to produce 1.116 moles of \( \mathrm{O}_2 \), we need\[ 1.116 \times 2 = 2.232 \text{ moles of } \mathrm{H}_2\mathrm{O}_2 \].

Convert Moles of H2O2 to Grams

To find the grams of \( \mathrm{H}_2\mathrm{O}_2 \), multiply the moles by its molar mass (34.02 g/mol):\[ 2.232 \times 34.02 = 75.901 \text{ grams of } \mathrm{H}_2\mathrm{O}_2 \].

Calculate Total Mass of Solution Needed

A \(3.00\%\) \( \mathrm{H}_2\mathrm{O}_2 \) solution means there are 3 grams of \( \mathrm{H}_2\mathrm{O}_2 \) per 100 grams of solution. To find the total mass of solution required:\[ \frac{75.901}{3.00/100} = 2530.033 \text{ grams of solution} \].

Key concepts

Stoichiometry

Stoichiometry is the concept that helps us understand the quantitative relationships in chemical reactions. When we look at a balanced chemical equation, stoichiometry tells us how much of each reactant is needed to form a particular amount of product.
For example, let's take the reaction given in the exercise:
\[ 2 \; \mathrm{H}_2\mathrm{O}_2 \rightarrow 2 \; \mathrm{H}_2\mathrm{O} + \mathrm{O}_2 \]From this equation, we see that to form one mole of oxygen gas \(\mathrm{O}_2 \), you need two moles of hydrogen peroxide \(\mathrm{H}_2\mathrm{O}_2 \).
Stoichiometry helps us convert from moles of one substance in the reaction to moles of any other substance, using the coefficients in the balanced equation.

• We know the moles of the product \( \mathrm{O}_2 \), which is 1.116 moles, as calculated from the given mass and molar mass.
• Using stoichiometry, we calculate that we need twice as many moles of \( \mathrm{H}_2\mathrm{O}_2 \), resulting in 2.232 moles.

Understanding these ratios is key in solving problems involving chemical reactions.

Solution Concentration

Solution concentration tells us how much solute is present in a given amount of solution. It can be expressed in various ways such as molarity, mass percent, etc. In this exercise, we are working with a 3% \( \mathrm{H}_2\mathrm{O}_2 \) solution.
"3% \( \mathrm{H}_2\mathrm{O}_2 \) solution" implies that there are 3 grams of hydrogen peroxide per every 100 grams of the solution.

• If you need to find out how much solution is required to have a specific amount of \( \mathrm{H}_2\mathrm{O}_2 \), you can use the mass percent as a conversion factor.
• In our problem, we calculated that 75.901 grams of \( \mathrm{H}_2\mathrm{O}_2 \) are needed. Since it's 3% of the solution, the total mass of the solution is determined by the equation: \( \frac{75.901}{3/100} = 2530.033 \) grams.

This skill of converting between mass, concentration, and total solution is very useful in the lab and real-world applications.

Balanced Chemical Equation

A balanced chemical equation represents a chemical reaction and shows the relation between reactants and products. The coefficients in the balanced equation ensure that the law of conservation of mass is not violated; that is, mass is neither created nor destroyed in a reaction.
In the given reaction:\[ 2 \; \mathrm{H}_2\mathrm{O}_2 \rightarrow 2 \; \mathrm{H}_2\mathrm{O} + \mathrm{O}_2 \]each molecule of oxygen gas formed comes from two molecules of hydrogen peroxide.

• Balancing the equation involves ensuring there is the same number of each type of atom on both sides of the equation. For instance, there are four hydrogen atoms and four oxygen atoms on both sides of the equation.
• This also allows us to use the equation for stoichiometry, turning the equation into a conversion tool between molecules, moles, and grams.

For any chemical reaction calculation, always make sure the chemical equation is balanced first.