91Ó°ÊÓ

Chemical reactions are processes where reactants are transformed into products. In this particular reaction, aluminum ore is converted into cryolite (\(\mathrm{Na_3AlF_6}\)) by combining various substances. The reactants here are hydrofluoric acid (\(\mathrm{HF}\)) and sodium aluminate (\(\mathrm{NaAlO_2}\)). By understanding this, we see a chemical equation as a symbolic representation of a chemical reaction. It shows the substances involved and their proportions.

In the provided equation:

• The left side contains the reactants: \(6\ \mathrm{HF}(g)\) and \(3\ \mathrm{NaAlO_2}(s)\).
• The right side shows the products formed: \(\mathrm{Na_3AlF_6}(s)\), \(3\ \mathrm{H_2O}(\ell)\), and \(\mathrm{Al_2O_3}(s)\).

This equation indicates that specific amounts of reactants yield specific amounts of products. The equation must be balanced to obey the law of conservation of mass, which states that atoms cannot be created or destroyed in chemical reactions. Understanding chemical reactions allows us to predict the outcomes when different substances interact.

Molar mass is crucial in stoichiometry for converting between mass and moles. Each atom has a unique molar mass, corresponding to its atomic weight found on the periodic table. The molar mass of a compound is the sum of the molar masses of its constituent elements. This is essential in converting measured gross mass of a substance into moles, which are more useful for stoichiometric calculations.

To find the molar mass of \(\mathrm{Na_3AlF_6}\) used in this reaction:

• Molar mass of \(\mathrm{Na} = 23.00 \ \text{g/mol}\)
• Molar mass of \(\mathrm{Al} = 26.98 \ \text{g/mol}\)
• Molar mass of \(\mathrm{F} = 19.00 \ \text{g/mol}\)

Therefore, the molar mass of \(\mathrm{Na_3AlF_6}\) is calculated as\[3 \times 23.00 + 26.98 + 6 \times 19.00 = 209.98 \ \text{g/mol}.\]This allows us to convert the 1,000 grams of \(\mathrm{Na_3AlF_6}\), as given in the problem, to moles using the formula: \(\text{moles of } \mathrm{Na_3AlF_6} = \frac{\text{mass of } \mathrm{Na_3AlF_6}}{\text{molar mass of } \mathrm{Na_3AlF_6}}\).

Balanced equations are critical in chemical reactions as they ensure that the same number of each type of atom is present on both sides of the equation. This balance is fundamental because it corresponds to the conservation of mass. Without a balanced equation, it would be impossible to correctly relate amounts of reactants to amounts of products.

In our specific equation:

• Reactants: \(6\ \mathrm{HF}\) and \(3\ \mathrm{NaAlO_2}\).
• Products: \(\mathrm{Na_3AlF_6}\), \(3\ \mathrm{H_2O}\), and \(\mathrm{Al_2O_3}\).

Each element is accounted for on both sides of the reaction. For example, there are six fluorine atoms both before and after the reaction due to the combination of \(6\ \mathrm{HF}\) forming \(\mathrm{Na_3AlF_6}\). The equation is balanced with respect to every element involved.

This balance allows us to apply stoichiometry accurately. For instance, knowing three moles of \(\mathrm{NaAlO_2}\) are needed per mole of \(\mathrm{Na_3AlF_6}\) is derived from a balance of all atomic ultimately helping us calculate the amount of reactants like \(\mathrm{NaAlO_2}\) needed. This becomes a powerful tool in planning chemical reactions in both laboratory and industrial settings.