Problem 98 - Chemistry and Chemical Reactivity

Question

Commercial sodium "hydrosulfite" is \(90.1 \%\) \(\mathrm{Na}_{2} \mathrm{S}_{2} \mathrm{O}_{4} .\) The sequence of reactions used to prepare the compound is $$\mathrm{Zn}(\mathrm{s})+2 \mathrm{SO}_{2}(\mathrm{g}) \rightarrow \mathrm{ZnS}_{2} \mathrm{O}_{4}(\mathrm{s})$$ $$\mathrm{ZnS}_{2} \mathrm{O}_{4}(\mathrm{s})+\mathrm{Na}_{2} \mathrm{CO}_{3}(\mathrm{aq}) \rightarrow \mathrm{ZnCO}_{3}(\mathrm{s})+\mathrm{Na}_{2} \mathrm{S}_{2} \mathrm{O}_{4}(\mathrm{aq})$$ (a) What mass of pure \(\mathrm{Na}_{2} \mathrm{S}_{2} \mathrm{O}_{4}\) can be prepared from \(125 \mathrm{kg}\) of \(\mathrm{Zn}, 500 .\) g of \(\mathrm{SO}_{2},\) and an excess of \(\mathrm{Na}_{2} \mathrm{CO}_{3} ?\) (b) What mass of the commercial product would contain the \(\mathrm{Na}_{2} \mathrm{S}_{2} \mathrm{O}_{4}\) produced using the amounts of reactants in part (a)?

Step-by-Step Solution

Answer

678.15 g of pure Na2S2O4 and 752.11 g of commercial product.

Step 1: Determine Molar Masses

First, we need to find the molar masses of the elements and compounds involved. - Zn (Zinc): \( 65.38 \, \text{g/mol} \)- \( \text{SO}_2 \): \( 32.07 + 16.00 \times 2 = 64.07 \, \text{g/mol} \)- \( \text{Na}_2\text{S}_2\text{O}_4 \): \((2 \times 22.99) + (2 \times 32.07) + (4 \times 16.00) = 174.14 \, \text{g/mol} \)

Step 2: Balanced Equation and Mole Ratios

Given the balanced equations: 1. \( \text{Zn} + 2\text{SO}_2 \rightarrow \text{ZnS}_2\text{O}_4 \)2. \( \text{ZnS}_2\text{O}_4 + \text{Na}_2\text{CO}_3 \rightarrow \text{ZnCO}_3 + \text{Na}_2\text{S}_2\text{O}_4\)From the first equation, 1 mole of \( \text{Zn} \) reacts with 2 moles of \( \text{SO}_2 \). From the second, 1 mole of \( \text{ZnS}_2\text{O}_4 \) produces 1 mole of \( \text{Na}_2\text{S}_2\text{O}_4 \).

Step 3: Calculate Moles of Reactants

Convert the masses to moles:- For \( 125 \, \text{kg Zn} = 125,000 \, \text{g Zn} \): \[ \frac{125,000 \, \text{g}}{65.38 \, \text{g/mol}} = 1910.8 \, \text{mol Zn} \] - For \( 500 \, \text{g SO}_2 \): \[ \frac{500 \, \text{g}}{64.07 \, \text{g/mol}} = 7.80 \, \text{mol SO}_2 \]

Step 4: Identify Limiting Reactant

The limiting reactant will determine the amount of \( \text{Na}_2\text{S}_2\text{O}_4 \) formed. For every mole of \( \text{Zn} \), 2 moles of \( \text{SO}_2 \) are required. Calculate the maximum \( \text{SO}_2 \) that can react with \( \text{Zn} \): \[ 1910.8 \, \text{mol Zn} \times 2 = 3821.6 \, \text{mol SO}_2 \] Since only 7.80 moles of \( \text{SO}_2 \) are available, \( \text{SO}_2 \) is the limiting reactant.

Step 5: Calculate Moles of Product

Given that \( \text{SO}_2 \) is the limiting reactant, calculate moles of \( \text{Na}_2\text{S}_2\text{O}_4 \): Since 2 moles of \( \text{SO}_2 \) produce 1 mole of \( \text{Na}_2\text{S}_2\text{O}_4 \), \[ \frac{7.80 \, \text{mol SO}_2}{2} = 3.90 \, \text{mol Na}_2\text{S}_2\text{O}_4 \]

Step 6: Convert Moles to Mass

Convert moles of \( \text{Na}_2\text{S}_2\text{O}_4 \) to mass:\[ 3.90 \, \text{mol} \times 174.14 \, \text{g/mol} = 678.15 \, \text{g} \text{Na}_2\text{S}_2\text{O}_4 \]

Step 7: Calculate Commercial Product Mass

Since the commercial product is 90.1% pure, calculate the mass of the commercial product:\[ \frac{678.15 \, \text{g}}{0.901} = 752.11 \, \text{g} \]

Key Concepts

Limiting Reactant

In the realm of stoichiometry, the concept of a limiting reactant plays a crucial role. When two or more reactants are involved in a chemical reaction, the reactant that is completely consumed first is known as the limiting reactant. This reactant limits the amount of product that can be formed during the reaction, effectively determining the maximum yield. In the given exercise, we had zinc ( Zn ) and sulfur dioxide ( SO_2 ) reacting to form a product.
The stoichiometry of the first reaction, as provided, reveals that one mole of zinc requires two moles of sulfur dioxide to fully react. Upon calculating, it turns out that sulfur dioxide, with only 7.80 moles available, becomes the limiting reactant because zinc could have reacted with as many as 3821.6 moles of SO_2 (if available). Hence, sulfur dioxide, being the limiting factor, controls the reaction's progress and consequently the amount of Na_2S_2O_4 formed.

Molar Mass

Molar mass is an essential concept in stoichiometry that refers to the mass of one mole of a substance. It is usually expressed in grams per mole ( g/mol ) and is determined by summing up the atomic masses of all the atoms present in a chemical formula. Molar mass helps convert atoms into grams, enabling us to relate macroscopic amounts of substances to their chemical equations.
In practice, as seen in the solution, we calculated the molar masses of the substances involved in the reactions. For example:

• Zinc ( Zn ): 65.38 g/mol
• Sulfur dioxide ( SO_2 ): 64.07 g/mol
• And Na_2S_2O_4 : 174.14 g/mol

Calculating the molar mass is an essential step, as it allows conversion from given masses to moles, and vice versa, paving the way for stoichiometric calculations involving the limiting reactant and product formation.

Chemical Reactions

Chemical reactions represent the process by which substances, known as reactants, are transformed into different substances, called products. These reactions adhere to the law of conservation of mass, meaning the total mass of reactants equals the total mass of products, provided the system is closed. In the exercise, the key reactions involve zinc and sulfur dioxide reacting to form ZnS_2O_4 , and subsequently producing Na_2S_2O_4 in the presence of Na_2CO_3 .
Balanced chemical equations are fundamental for understanding these reactions because they depict the relationship between the reactants and products in terms of moles. For instance:

• Zn + 2 SO_2 → ZnS_2O_4
• ZnS_2O_4 + Na_2CO_3 → ZnCO_3 + Na_2S_2O_4

These equations tell us how molecules interact and in what proportions they react, which is vital for further stoichiometric calculations.

Mass to Moles Conversion

Mass to moles conversion is a pivotal aspect of stoichiometry that enables calculation from a given mass to the amount of substance in moles. This conversion is possible using the formula:
Number of moles = mass (g) / molar mass (g/mol).
For instance, to find out how many moles of zinc react, we take the given mass of zinc and divide by its molar mass:\[\frac{125,000 \, \text{g Zn}}{65.38 \, \text{g/mol}} = 1910.8 \, \text{mol Zn}\] Similarly, converting 500 grams of sulfur dioxide to moles:\[\frac{500 \, \text{g SO}_2}{64.07 \, \text{g/mol}} = 7.80 \, \text{mol SO}_2\]These conversions are cornerstone operations in solving stoichiometric problems, forming the basis for determining limiting reactants and calculating expected product yields.