Question

If sodium peroxide is added to water, elemental 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 will be \(1.33\, g\).

Step-by-step solution

Find the moles of sodium peroxide

First, look up the molar mass of sodium peroxide (\(\mathrm{Na}_{2}\mathrm{O}_{2}\)): Molar mass of \(\mathrm{Na} = 22.99\, g/mol\), Molar mass of \(\mathrm{O} = 16.00\, g/mol\), Therefore, Molar mass of \(\mathrm{Na}_{2}\mathrm{O}_{2} = 2 \times 22.99 + 2 \times 16.00 = 77.98\, g/mol\). Now, we can convert the mass of sodium peroxide to moles: \( \text{moles of }\mathrm{Na}_{2}\mathrm{O}_{2} = \frac{\text{mass}}{\text{molar mass}} = \frac{3.25\,g}{77.98\, g/mol} = 0.0417\, \text{moles} \)

Stoichiometry to determine moles of oxygen gas

The balanced equation is given as: \(\mathrm{Na}_{2}\mathrm{O}_{2}(s) + \mathrm{H}_{2}\mathrm{O}(l) \rightarrow \mathrm{NaOH}(aq) + \mathrm{O}_{2}(g)\) From the balanced chemical equation, 1 mole of \(\mathrm{Na}_{2}\mathrm{O}_{2}\) produces 1 mole of \(\mathrm{O}_{2}(g)\). Thus, using stoichiometry, we can determine the moles of \(\mathrm{O}_{2}(g)\) produced: \( \text{moles of }\mathrm{O}_{2}(g) = \text{moles of } \mathrm{Na}_{2}\mathrm{O}_{2} = 0.0417\, \text{moles} \)

Calculate the mass of oxygen gas

Now we need to convert the moles of \(\mathrm{O}_{2}(g)\) back to mass. The molar mass of \(\mathrm{O}_{2}(g)\) is: Molar mass of \(\mathrm{O}_{2} = 2 \times 16.00 = 32.00\, g/mol\). To calculate the mass of \(\mathrm{O}_{2}(g)\) produced, simply multiply the moles by the molar mass: \( \text{mass of }\mathrm{O}_{2}(g) = \text{moles} \times \text{molar mass} = 0.0417\, \text{moles} \times 32.00\, \mathrm{g/mol} = 1.33\,g \) Therefore, the mass of oxygen gas produced will be \(1.33\, g\).

Key concepts

Chemical Reactions

In chemistry, a chemical reaction is a process where reactants are transformed into products. This involves the breaking of old chemical bonds and the formation of new ones.
A classic example is the reaction between sodium peroxide and water, where sodium peroxide reacts with water to produce sodium hydroxide and oxygen gas.

• Reactants: These are the starting materials in a chemical reaction—in this case, sodium peroxide (\(\mathrm{Na}_{2} \mathrm{O}_{2}\)) and water (\(\mathrm{H}_{2} \mathrm{O}\)).
• Products: These are the new substances formed, such as sodium hydroxide (\(\mathrm{NaOH}\)) and oxygen gas (\(\mathrm{O}_{2}\)).

Chemical reactions are fundamental in chemistry because they allow us to understand how substances interact and transform.
They are everywhere—from the digestion of food to combustion in cars.

Molar Mass Calculation

Molar mass is the mass of a given substance divided by the amount of substance. It is important for converting between the mass of a chemical and the moles of a chemical substance.
To find the molar mass of a compound, you:

• List each element in the compound.
• Multiply the atomic mass of each element by the number of times it appears (its subscript in the formula).
• Add these values together.

For example, the molar mass of sodium peroxide, \(\mathrm{Na}_{2}\mathrm{O}_{2}\), is calculated by adding twice the atomic mass of sodium (\(22.99\, \mathrm{g/mol}\)) and twice the atomic mass of oxygen (\(16.00\, \mathrm{g/mol}\)):
\[ 2 \times 22.99 + 2 \times 16.00 = 77.98 \, \mathrm{g/mol} \] Understanding molar mass is crucial for stoichiometry, as it allows you to convert a mass of a substance to the number of moles, which is necessary for predicting how much product you can get from a reaction.

Balanced Chemical Equation

A balanced chemical equation accurately represents the reactants and products in a chemical reaction with their correct stoichiometric coefficients.
This ensures that the law of conservation of mass is followed, meaning matter is neither created nor destroyed.

• Coefficients: These numbers in front of compounds ensure that the number of each type of atom is the same in both reactants and products.
• For the reaction of sodium peroxide with water, the equation was already balanced: \[ \mathrm{Na}_{2}\mathrm{O}_{2}(s) + \mathrm{H}_{2}\mathrm{O}(l) \rightarrow \mathrm{NaOH}(aq) + \mathrm{O}_{2}(g) \]

Without a balanced equation, it would be impossible to determine the correct amount of products formed.
This is critical in stoichiometric calculations, which rely on these ratios to predict the quantity of products from given reactants.