Question

It has been found that the following sequence can be used to prepare sodium sulfate, \(\mathrm{Na}_{2} \mathrm{SO}_{4}\) : \(\mathrm{S}(\mathrm{s})+\mathrm{O}_{2}(\mathrm{~g}) \rightarrow \mathrm{SO}_{2}(\mathrm{~g})\) \(2 \mathrm{SO}_{2}(\mathrm{~g})+\mathrm{O}_{2}(\mathrm{~g}) \rightarrow 2 \mathrm{SO}_{3}(\mathrm{~g})\) \(\mathrm{SO}_{3}(\mathrm{~g})+\mathrm{H}_{2} \mathrm{O}(\ell) \rightarrow \mathrm{H}_{2} \mathrm{SO}_{4}(\ell)\) \(2 \mathrm{NaOH}+\mathrm{H}_{2} \mathrm{SO}_{4} \rightarrow \mathrm{Na}_{2} \mathrm{SO}_{4}+2 \mathrm{H}_{2} \mathrm{O}\) If you performed this sequence of reactions, how many moles of \(\mathrm{Na}_{2} \mathrm{SO}_{4}\) could possibly be produced if you start with 1 mole of sulfur? How many moles are possible with \(4.5\) gof water?

Short answer

The maximum moles of \(\mathrm{Na}_{2}\mathrm{SO}_{4}\) that could possibly be produced are 0.25 moles, when comparing the sequences starting with 1 mole of sulfur and \(4.5\,\text{g}\) of water.

Step-by-step solution

Calculate moles of \(\mathrm{SO}_{2}\) from sulfur

Since the first reaction has a stoichiometric ratio of 1:1 between sulfur and \(\mathrm{SO}_{2}\), we have 1 mole of \(\mathrm{SO}_{2}\) produced.

Calculate moles of \(\mathrm{SO}_{3}\) from \(\mathrm{SO}_{2}\)

The second reaction has a stoichiometric ratio of 2:1 for \(\mathrm{SO}_{2}\) to \(\mathrm{O}_{2}\). From 1 mole of \(\mathrm{SO}_{2}\), we calculate the moles of \(\mathrm{SO}_{3}\) as follows: (\(\frac{1}{2}\)) x 1 mole = 0.5 moles of \(\mathrm{SO}_{3}\) produced.

Calculate moles of \(\mathrm{H}_{2}\mathrm{SO}_{4}\) from \(\mathrm{SO}_{3}\)

The third reaction has a 1:1 stoichiometric ratio for \(\mathrm{SO}_{3}\) to \(\mathrm{H}_{2}\mathrm{SO}_{4}\). So, from 0.5 moles of \(\mathrm{SO}_{3}\), we will get 0.5 moles of \(\mathrm{H}_{2}\mathrm{SO}_{4}\).

Calculate moles of \(\mathrm{Na}_{2}\mathrm{SO}_{4}\) from \(\mathrm{H}_{2}\mathrm{SO}_{4}\)

In the fourth reaction, the stoichiometric ratio of \(\mathrm{H}_{2}\mathrm{SO}_{4}\) to \(\mathrm{Na}_{2}\mathrm{SO}_{4}\) is 1:1. So, from 0.5 moles of \(\mathrm{H}_{2}\mathrm{SO}_{4}\), we will get 0.5 moles of \(\mathrm{Na}_{2}\mathrm{SO}_{4}\). 2. For the second sequence starting with \(4.5 \mathrm{~g}\) of water:

Calculate moles of water

Molar mass of water(\(\mathrm{H}_{2}\mathrm{O}\)) = \(18.02 \mathrm{~g/mol}\). Since moles = \(\frac{\text{mass}}{\text{molar mass}}\), we get: Moles of water = (\(\frac{4.5\,\text{g}}{18.02\,\text{g/mol}}\))= 0.25 moles

Calculate moles of \(\mathrm{H}_{2}\mathrm{SO}_{4}\) from water

In the third reaction, the stoichiometric ratio of \(\mathrm{SO}_{3}\) to water is 1:1. Since we have ample moles of \(\mathrm{SO}_{3}\) from Step 2, we have: Moles of \(\mathrm{H}_{2}\mathrm{SO}_{4}\) produced from 0.25 moles of water = 0.25 moles

Calculate moles of \(\mathrm{Na}_{2}\mathrm{SO}_{4}\) from \(\mathrm{H}_{2}\mathrm{SO}_{4}\)

In the fourth reaction, the stoichiometric ratio of \(\mathrm{H}_{2}\mathrm{SO}_{4}\) to \(\mathrm{Na}_{2}\mathrm{SO}_{4}\) is 1:1. So, from 0.25 moles of \(\mathrm{H}_{2}\mathrm{SO}_{4}\), we will get 0.25 moles of \(\mathrm{Na}_{2}\mathrm{SO}_{4}\).

Determine the actual moles of \(\mathrm{Na}_{2}\mathrm{SO}_{4}\) produced

Since \(\mathrm{Na}_{2}\mathrm{SO}_{4}\) is produced in both sequences, the limiting reagent will be the one with the fewer moles. Comparing the moles from both sequences, 0.5 moles (from 1 mole of sulfur) and 0.25 moles (from \(4.5\,\text{g}\) of water), we find the limiting reagent to be the sequence starting with \(4.5\,\text{g}\) of water. Therefore, the maximum moles of \(\mathrm{Na}_{2}\mathrm{SO}_{4}\) that could possibly be produced are 0.25 moles.

Key concepts

Chemical Reactions

Chemical reactions are processes that transform one or more substances into different substances. This occurs through breaking and forming bonds between atoms. Each reaction involves reactants (starting materials) and products (newly formed substances).
In the case of sodium sulfate synthesis, the process involves a series of four distinct reactions. Each step transforms certain reactants into products until sodium sulfate (\(\text{Na}_2\text{SO}_4\)) is produced.

• The first reaction combines sulfur (\(\text{S}\)) with oxygen (\(\text{O}_2\)) to form sulfur dioxide (\(\text{SO}_2\))
• The second reaction further oxidizes sulfur dioxide to sulfur trioxide (\(\text{SO}_3\))
• Then, sulfur trioxide reacts with water (\(\text{H}_2\text{O}\)) to produce sulfuric acid (\(\text{H}_2\text{SO}_4\))
• Finally, sulfuric acid reacts with sodium hydroxide (\(\text{NaOH}\)) to form sodium sulfate and water.

By understanding how these reactions occur step-by-step, you can see the transformation of simple reactants into a valuable product like sodium sulfate.

Mole Calculations

Mole calculations are essential in chemistry to determine the amount of substances involved in a chemical reaction. A "mole" is a unit that measures a substance's amount, allowing chemists to count atoms and molecules in terms of grams.

To carry out mole calculations, you need to know the molar mass of the involved substances, which is the mass of one mole of that substance. For instance, the molar mass of water (\(\text{H}_2\text{O}\)) is 18.02 grams per mole. By using the formula \[ \text{moles} = \frac{\text{mass}}{\text{molar mass}} \] you can calculate how many moles are present in a given mass.

• In this example, the calculation starts with converting the mass of water (4.5 grams) to moles, resulting in 0.25 moles of water.
• We used a similar approach for all reactants and products across the reactions.

These calculations are pivotal in predicting the amounts of reactants required or products formed during a chemical reaction.

Limiting Reagent

The limiting reagent (or reactant) in a chemical reaction is the substance that is completely consumed first, limiting the amount of product that can be formed. It is vital to identify the limiting reagent to predict how much product will result from a reaction.

In the synthesis of sodium sulfate, both sulfur and water were used as starting materials, but their amounts were different:

• 1 mole of sulfur produced 0.5 moles of sodium sulfate.
• 4.5 grams of water produced 0.25 moles of sodium sulfate.

When comparing these results, the sequence with water becomes the limiting factor, as it leads to fewer moles of product. Identifying the limiting reagent involves comparing the moles of possible product from each reactant. Whichever reactant yields the fewer moles of product becomes the limiting reactant. Understanding this concept ensures that resources are used efficiently and helps avoid unnecessary waste in chemical processes.

Sodium Sulfate Synthesis

Sodium sulfate synthesis involves a multi-step reaction sequence to produce a commonly used industrial compound. Each step is designed to change reactants into intermediates and finally into sodium sulfate.

The process complexity results from needing to manage different starting materials, reactant ratios, and reaction conditions.

• Initially, the synthesis produces sulfur dioxide through oxidation of sulfur, which is further oxidized to sulfur trioxide.
• This compound then reacts with water to produce sulfuric acid, a crucial intermediate.
• Completing the synthesis, sulfuric acid reacts with sodium hydroxide to produce sodium sulfate and water.

Nailing each reaction's stoichiometry and understanding how each step connects is key. Sodium sulfate is crucial in many industries, including detergents, paper manufacturing, and textile production, highlighting why mastering its synthesis is valuable for chemists and industrial processes alike.