Problem 35 - Chemistry: An Atoms-Focused Approach

Question

One step in the conversion of aluminum ore into aluminum metal involves the synthesis of cryolite \(\left(\mathrm{Na}_{3} \mathrm{AlF}_{6}\right)\) in the following reaction: $$6 \mathrm{HF}(g)+3 \mathrm{NaAlO}_{2}(s) \rightarrow \mathrm{Na}_{3} \mathrm{AlF}_{6}(s)+3 \mathrm{H}_{2} \mathrm{O}(\ell)+\mathrm{Al}_{2} \mathrm{O}_{3}(s)$$ How much NaAIO \(_{2}\) (sodium aluminate) is required to produce \(1.00 \mathrm{kg}\) of \(\mathrm{Na}_{3} \mathrm{AlF}_{6} ?\)

Step-by-Step Solution

Answer

Based on the given information, determine the mass of NaAlO2 required to produce 1.00 kg of Na3AlF6. 1. Write down the balanced chemical equation: 6 HF(g) + 3 NaAlO2(s) → Na3AlF6(s) + 3 H2O(l) + Al2O3(s) 2. Calculate the moles of Na3AlF6 produced (use the molar mass): Moles of Na3AlF6 = mass of Na3AlF6 / molar mass of Na3AlF6 3. Calculate the moles of NaAlO2 required: Moles of NaAlO2 required = 3 × moles of Na3AlF6 4. Convert moles of NaAlO2 to grams (use the molar mass): Mass of NaAlO2 required = moles of NaAlO2 × molar mass of NaAlO2 Using the appropriate molar masses, calculate the mass of NaAlO2 required to produce 1.00 kg of Na3AlF6.

Step 1: Write down the balanced chemical equation

The given balanced chemical equation is: $$6 \mathrm{HF}(g)+3 \mathrm{NaAlO}_{2}(s) \rightarrow \mathrm{Na}_{3}\mathrm{AlF}_{6}(s)+3 \mathrm{H}_{2} \mathrm{O}(\ell)+\mathrm{Al}_{2}\mathrm{O}_{3}(s)$$

Step 2: Calculate the moles of Na\(_3\)AlF\(_6\) produced

We are given that 1.00 kg (1000 g) of Na\(_3\)AlF\(_6\) is produced. To find the moles of Na\(_3\)AlF\(_6\), we can use its molar mass: Molar mass of Na\(_3\)AlF\(_6 = 3 \times (\text{molar mass of Na}) + (\text{molar mass of Al}) + 6 \times (\text{molar mass of F})\) Therefore, moles of Na\(_3\)AlF\(_6\) = \(\frac{\text{mass of Na}_{3}\text{AlF}_{6}}{\text{molar mass of Na}_{3}\text{AlF}_{6}}\)

Step 3: Calculate the moles of NaAlO\(_2\) required

Using stoichiometry, we can determine the moles of NaAlO\(_2\) required for the production of the given moles of Na\(_3\)AlF\(_6\). From the balanced chemical equation, 3 moles of NaAlO\(_2\) are needed to produce 1 mole of Na\(_3\)AlF\(_6\). Moles of NaAlO\(_2\) required = \(3 \times (\text{moles of Na}_{3}\text{AlF}_{6})\)

Step 4: Convert moles of NaAlO\(_2\) to grams

Now that we have the moles of NaAlO\(_2\) required, we can convert this value to grams using its molar mass: Molar mass of NaAlO\(_2\) = \((\text{molar mass of Na}) + (\text{molar mass of Al}) + 2 \times (\text{molar mass of O})\) Mass of NaAlO\(_2\) required = (moles of NaAlO\(_2\)) \(\times\) (molar mass of NaAlO\(_2\)) By following the above steps and using the appropriate molar masses, the mass of NaAlO\(_2\) required to produce 1.00 kg of Na\(_3\)AlF\(_6\) can be calculated.

Key Concepts

Chemical reactions

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 calculation

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

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.