Question

When active metals such as sodium are exposed to air, they quickly form a coating of metal oxide. The balanced equation for the reaction of sodium metal with oxygen gas is $$ 4 \mathrm{Na}(s)+\mathrm{O}_{2}(g) \longrightarrow 2 \mathrm{Na}_{2} \mathrm{O}(s) $$ Suppose a piece of sodium metal gains \(2.05 \mathrm{~g}\) of mass after being exposed to air. Assume that this gain can be attributed to its reaction with oxygen. (a) What mass of \(\mathrm{O}_{2}\) reacted with the \(\mathrm{Na}\) ? (b) What mass of Na reacted? (c) What mass of \(\mathrm{Na}_{2} \mathrm{O}\) formed?

Short answer

(a) The mass of O2 that reacted with Na is 2.05 g. (b) The mass of Na that reacted is 11.8025 g. (c) The mass of Na2O formed is 15.8875 g.

Step-by-step solution

Determine moles of oxygen added

Assuming that the weight gain of 2.05 g is due to the reaction with oxygen, we know that this is the weight of the oxygen. Oxygen has a molar mass of 16g/mol, so there are 2.05g / 16g/mol = 0.128125 moles of oxygen.

Use stoichiometry to find moles of Na and Na2O

From the balanced chemical equation, we know that 4 moles of Na react with 1 mole of O2 to produce 2 moles of Na2O. Therefore, (0.128125 moles O2) * (4 moles Na / 1 mole O2) = 0.5125 moles of Na reacted. Also, (0.128125 moles O2) * (2 moles Na2O / 1 mole O2) = 0.25625 moles of Na2O formed.

Convert moles back to grams to find the mass

The molar mass of Na is 23 g/mol and the molar mass of Na2O is 62 g/mol. Therefore, (0.5125 moles Na) * (23 g Na / 1 mole Na) = 11.8025 g Na reacted. (0.25625 moles Na2O) * (62 g Na2O / 1 mole Na2O) = 15.8875 g Na2O formed.

Key concepts

Chemical Reaction Balancing

In any chemical reaction, atoms are neither created nor destroyed, and thus the same number of atoms must exist on both the reactant and the product sides. This is the core principle behind balancing chemical equations. For example, when sodium (\textbf{Na}) reacts with oxygen (\textbf{O}\textsubscript{2}), the product is sodium oxide (\textbf{Na}\textsubscript{2}\textbf{O}). The balanced reaction is shown as: \[4\text{Na}(s) + \text{O}_2(g) \rightarrow 2\text{Na}_2\text{O}(s)\].
The coefficients indicate the ratio of moles required for the reaction to occur. Here, four moles of sodium are needed to react with one mole of diatomic oxygen to produce two moles of sodium oxide. Understanding and applying this precise stoichiometric ratio is essential for accurately determining the amounts of reactants and products involved in a chemical reaction.

Molar Mass Calculation

The molar mass is the weight of one mole of a substance, typically expressed in grams per mole (g/mol). It is calculated by adding up the atomic masses of all atoms in the molecule, which are found on the periodic table. For instance, oxygen gas (\textbf{O}\textsubscript{2}) consists of two oxygen atoms, and since each oxygen atom has an atomic mass of approximately 16 units (g/mol), the molar mass of oxygen gas becomes \[(2 \times 16) \text{ g/mol} = 32 \text{ g/mol}\].
This calculation is essential in converting between mass and moles of a substance. In the example given in the exercise, we used the molar mass of \textbf{O}\textsubscript{2} to determine that a 2.05 g gain in mass corresponds to \[\frac{2.05 \, \text{g}}{32 \, \text{g/mol}} = 0.128125 \, \text{moles}\] of \textbf{O}\textsubscript{2}, which is the starting point for finding out the amounts of reactants and products.

Mass-Mole Conversion

In stoichiometry, conversions between mass and moles are frequent and crucial for quantifying the amount of reactants and products. We use the molar mass as a conversion factor to change grams into moles or vice versa. Taking sodium (\textbf{Na}) as an example, to find how many grams are in 0.5125 moles of sodium, we calculate \[(0.5125 \text{ moles Na}) \times (23 \text{ g Na/mol}) = 11.8025 \text{ g Na}\].
Likewise, to find out how many grams of sodium oxide (\textbf{Na}\textsubscript{2}\textbf{O}) are formed from 0.25625 moles, we multiply the moles by the molar mass of \textbf{Na}\textsubscript{2}\textbf{O}: \[(0.25625 \text{ moles Na}\textsubscript{2}\text{O}) \times (62 \text{ g Na}\textsubscript{2}\text{O/mol}) = 15.8875 \text{ g Na}\textsubscript{2}\text{O}\].
By performing these mass-mole conversions, we can connect the theoretical world of chemical equations to the practical, measurable laboratory quantities.