91Ó°ÊÓ

A balanced chemical equation is essential in stoichiometry. It ensures that the number of atoms for each element is the same on both sides of the reaction. This follows the law of conservation of mass. For our given chemical reaction:

4 Na(s) + O2(g) → 2 Na2O(s),

we see that 4 moles of Sodium (Na) react with 1 mole of Oxygen (O2) to produce 2 moles of Sodium Oxide (Na2O).

Understanding this ratio is crucial to solving stoichiometric problems. This ensures that we correctly relate the amounts of reactants and products.

Molar mass is the mass of one mole of a substance, typically expressed in grams per mole (g/mol). This concept is fundamental when converting between the mass of a substance and the moles of a substance.

For example, the molar mass of Sodium (Na) is approximately 22.99 g/mol, whereas the molar mass of Sodium Oxide (Na2O) is approximately 61.98 g/mol. Oxygen (O2), being a diatomic molecule, has a molar mass of approximately 32.00 g/mol.

These values are used to convert given masses into moles for stoichiometric calculations. For instance, to find the moles of Na in 57.5 g, you would use:

\[ \text{Moles of Na} = \frac{57.5 \text{ g}}{22.99 \text{ g/mol}} \ \text{Moles of Na} \thickapprox 2.50 \text{ moles} \ \text{} \]

The mole-to-mole ratio from a balanced chemical equation allows us to relate the number of moles of reactants to the number of moles of products. In our example:

4 Na(s) + O2(g) → 2 Na2O(s),

the ratio between Sodium (Na) to Sodium Oxide (Na2O) is 4:2 or 2:1. This means that 4 moles of Na produce 2 moles of Na2O.

If we know that we have 2.50 moles of Na, we can calculate the moles of Na2O by setting up a ratio:
\[ \text{Moles of Na2O} = \frac{2}{4} \times 2.50 = 1.25 \text{ moles} \]

Similar calculations apply when determining the moles of Oxygen (O2) needed for specific amounts of reactants or products.

Mass calculations involve converting between grams and moles using the molar mass and then using the mole-to-mole ratio to find the required masses of other substances. Let's go through an example:

Given 57.5 g of Na, how many grams of Na2O are produced?

First, convert mass of Na to moles:

\[ \text{Moles of Na} = \frac{57.5 \text{ g}}{22.99 \text{ g/mol}} \ \text{Moles of Na} \thickapprox 2.50 \text{ moles} \]

Then, use the mole-to-mole ratio to find moles of Na2O:

\[ \text{Moles of Na2O} = \frac{2}{4} \times 2.50 \]
\[ \text{Moles of Na2O} = 1.25 \text{ moles} \]

Finally, convert moles of Na2O to grams using its molar mass:

\[ \text{Mass of Na2O} = 1.25 \text{ moles} \times 61.98 \text{ g/mol} \]
\[ \text{Mass of Na2O} = 77.48 \text{ g} \ \text{} \]

Exact steps can be followed to solve problems involving different substances and quantities.