Question

If sodium peroxide is added to water, clemental oxygen gas is generated: $$ \mathrm{Na}_{2} \mathrm{O}_{2}(s)+\mathrm{H}_{2} \mathrm{O}(l) \rightarrow \mathrm{NaOH}(a q)+\mathrm{O}_{2}(g) $$ Suppose \(3.25 \mathrm{~g}\) of sodium peroxide is added to a large excess of water. What mass of oxygen gas will be produced?

Short answer

The mass of oxygen gas produced when 3.25 g of sodium peroxide reacts with water is approximately 1.334 grams.

Step-by-step solution

Write down the balanced chemical equation

The balanced chemical equation is given as: \[ \mathrm{Na}_{2} \mathrm{O}_{2}(s)+\mathrm{H}_{2} \mathrm{O}(l) \rightarrow \mathrm{NaOH}(a q)+\mathrm{O}_{2}(g) \]

Calculate the moles of sodium peroxide

We are given that 3.25 grams of sodium peroxide are added to the water. To find the moles of Na2O2, we can use the following formula: \[n = \frac{m}{M}\] Where m = mass of Na2O2 (3.25 g) M = molar mass of Na2O2 (22.99 g/mol * 2 + 16.00 g/mol * 2 = 45.98 g/mol + 32.00 g/mol = 77.98 g/mol) n = moles of Na2O2 \[\begin{aligned} n = \frac{3.25 \ \text{g}}{77.98 \ \text{g/mol}} \ !=\approx 0.0417 \ \text{moles} \end{aligned}\]

Find the moles of oxygen gas produced

From the balanced chemical equation, we can see that one mole of Na2O2 reacts to produce one mole of O2 gas. So, 0.0417 moles of Na2O2 react to produce the same number (0.0417 moles) of O2 gas.

Calculate the mass of oxygen gas produced

Now, we can use the moles of oxygen gas and its molar mass to find the mass of O2 gas produced: \[m_{O_2} = n_{O_2} \times M_{O_2}\] Where m_O2 = mass of O2 n_O2 = moles of O2 (0.0417 moles) M_O2 = molar mass of O2 (16.00 g/mol * 2 = 32.00 g/mol) \[\begin{aligned} m_{O_{2}} = 0.0417 \ \text{moles} \times 32.00 \ \text{g/mol} = \approx 1.334 \ \text{g} \end{aligned}\]

Final Answer

The mass of oxygen gas produced when 3.25 g of sodium peroxide reacts with water is approximately 1.334 grams.

Key concepts

Stoichiometry

The chemistry students' compass, stoichiometry, guides us through the quantitative relationships within chemical reactions. It's like following a recipe; we start with specific amounts of ingredients (reactants) to end up with a set amount of product.

In our sodium peroxide and water reaction, stoichiometry becomes critical to predict the mass of oxygen gas formed. By knowing the exact ratio of reactants to products from the balanced equation, we can navigate from the initial mass of a reactant to the mass of the desired product.

Molar Mass

Molar mass, the weight of a chemical's one mole, is a fundamental bridge between the microscopic world of atoms and the scale we measure in labs. Think of it as the molecular equivalent of a 'price per pound' tag in a grocery store.

For sodium peroxide, we calculate its molar mass by summing the molar masses of sodium and oxygen, which helps us unlock the answer to 'how many moles are in 3.25 grams of sodium peroxide?' It's the conversion factor that helps us bake the perfect molecular cake every time!

Balanced Chemical Equation

At the heart of every chemistry problem lies this simple truth: matter is neither created nor destroyed, only transformed. This principle is represented by a chemical equation, a written shorthand for the reactions happening atom by atom.

For our exercise, a balanced chemical equation ensures that for each sodium peroxide molecule reacting, there's a precise amount of sodium hydroxide and oxygen gas produced, ensuring the universal law of conservation of mass is upheld, even in a simple classroom experiment.

Moles to Grams Conversion

The concept of 'moles to grams conversion' dances elegantly between the abstract world of atoms and our tangible reality. Our equation \( n = \frac{m}{M} \) is a converter, turning moles, a count of particles, into grams, something you can feel in your hand.

When we learn that 0.0417 moles of sodium peroxide will produce an equal number of moles of oxygen, this conversion becomes our essential tool to find out just how much oxygen, in grams, we can expect to capture when adding sodium peroxide to water.