Problem 71 - Chemistry and Chemical Reactivity

Question

Commercial sodium "hydrosulfite" is \(90.1 \%\) pure \(\mathrm{Na}_{2} \mathrm{S}_{2} \mathrm{O}_{4} .\) The sequence of reactions used to prepare the compound is $$\begin{aligned} \mathrm{Zn}(\mathrm{s})+2 \mathrm{SO}_{2}(\mathrm{g}) & \longrightarrow \mathrm{ZnS}_{2} \mathrm{O}_{4}(\mathrm{s}) \\ \mathrm{ZnS}_{2} \mathrm{O}_{4}(\mathrm{s})+\mathrm{Na}_{2} \mathrm{CO}_{3}(\mathrm{aq}) & \longrightarrow \mathrm{ZnCO}_{3}(\mathrm{s})+\mathrm{Na}_{2} \mathrm{S}_{2} \mathrm{O}_{4}(\mathrm{aq}) \end{aligned}$$ (a) What mass of pure \(\mathrm{Na}_{2} \mathrm{S}_{2} \mathrm{O}_{4}\) can be prepared from \(125 \mathrm{kg}\) of \(\mathrm{Zn}, 500 \mathrm{g}\) 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

1.36 kg of pure Na2S2O4 can be produced, requiring about 1.51 kg of the commercial product.

Step 1: Convert Mass of Zn to Moles

To start, calculate the moles of zinc using its molar mass: \[ \text{Molar mass of Zn} = 65.38 \, \text{g/mol} = 0.06538 \, \text{kg/mol} \] \[ \text{Moles of Zn} = \frac{\text{Mass of Zn}}{\text{Molar mass of Zn}} = \frac{125 \, \text{kg}}{0.06538 \, \text{kg/mol}} \approx 1911.95 \, \text{mol} \]

Step 2: Convert Mass of SO2 to Moles

Next, determine the moles of \( \mathrm{SO}_{2} \): \[ \text{Molar mass of SO}_{2} = 64.07 \, \text{g/mol} = 0.06407 \, \text{kg/mol} \] \[ \text{Moles of SO}_{2} = \frac{\text{Mass of SO}_{2}}{\text{Molar mass of SO}_{2}} = \frac{0.500 \, \text{kg}}{0.06407 \, \text{kg/mol}} \approx 7.80 \, \text{mol} \]

Step 3: Determine Limiting Reactant

The reaction \( \mathrm{Zn} + 2 \mathrm{SO}_{2} \longrightarrow \mathrm{ZnS}_{2} \mathrm{O}_{4} \) shows 1 mole of Zn reacts with 2 moles of SO2. Therefore, to react with 7.80 moles of SO2, we need: \[ \text{Required moles of Zn} = \frac{7.80}{2} = 3.90 \, \text{mol} \] Since 3.90 moles \( < 1911.95 \) moles of Zn available, SO2 is the limiting reactant.

Step 4: Calculate Moles of Na2S2O4 Produced

From the balanced equations, 1 mole of \( \mathrm{SO}_{2} \) produces 1 mole of \( \mathrm{Na}_{2} \mathrm{S}_{2} \mathrm{O}_{4} \). Therefore: \[ \text{Moles of Na}_{2} \text{S}_{2} \text{O}_{4} = 7.80 \, \text{mol} \]

Step 5: Convert Moles to Mass of Pure Na2S2O4

Using the molar mass of \( \mathrm{Na}_{2} \mathrm{S}_{2} \mathrm{O}_{4} \), calculate the mass of pure product: \[ \text{Molar mass of Na}_{2} \text{S}_{2} \text{O}_{4} = 174.1 \, \text{g/mol} = 0.1741 \, \text{kg/mol} \] \[ \text{Mass of pure Na}_{2} \text{S}_{2} \text{O}_{4} = 7.80 \, \text{mol} \times 0.1741 \, \text{kg/mol} \approx 1.36 \, \text{kg} \]

Step 6: Calculate Mass of Commercial Product

Given the purity is 90.1%, calculate the commercial mass: \[ 1.36 \, \text{kg of pure Na}_{2} \text{S}_{2} \text{O}_{4} = 0.901 \times \text{mass of commercial product} \] \[ \text{Mass of commercial product} = \frac{1.36}{0.901} \approx 1.51 \, \text{kg} \]

Key Concepts

Limiting Reactant

In chemical reactions, the limiting reactant is the substance that runs out first, stopping the reaction because there's nothing left for the other reactants to react with. To identify the limiting reactant, you need to compare the mole ratio of the reactants used to the mole ratio in the balanced chemical equation.
In this exercise, the equation is \( \text{Zn} + 2 \text{SO}_2 \rightarrow \text{ZnS}_2\text{O}_4 \). Here, 1 mole of zinc reacts with 2 moles of sulfur dioxide.

• We calculated 1911.95 moles of Zn and 7.80 moles of \( \text{SO}_2 \).
• To use all 7.80 moles of \( \text{SO}_2 \), we need 3.90 moles of Zn (calculated as \( \frac{7.80}{2} \)).
• Since 3.90 < 1911.95, \( \text{SO}_2 \) is the limiting reactant, meaning no more \( \text{Na}_2\text{S}_2\text{O}_4 \) can be produced once \( \text{SO}_2 \) is used up.

Molar Mass Calculation

Molar mass is key to converting between grams and moles. It's the mass of one mole of a chemical and is usually expressed in g/mol. To get the molar mass, sum the atomic masses of all the atoms in the molecule.
For example, let's calculate the molar masses needed in this problem:

• **Zinc (Zn):** 65.38 g/mol. This tells us how much 1 mole of Zn weighs.
• **Sulfur Dioxide (SO2):** 64.07 g/mol. Calculated by adding the atomic masses (S = 32.07, O = 16.00).
• **Sodium Hydrosulfite (Na2S2O4):** 174.1 g/mol. This comes from adding the atomic masses of all elements in the compound (Na = 22.99, S = 32.07, O = 16.00 x 4).

Knowing these allows you to convert given masses to moles, which is essential for understanding how much of each reactant you have and predicting how much product you can make.

Chemical Reactions

Chemical reactions involve rearranging atoms to transform reactants into products. Writing balanced equations is crucial since it clearly illustrates the reacting substances and their reactions.
In our exercise, the reactions are:

• \( \text{Zn} + 2 \text{SO}_2 \rightarrow \text{ZnS}_2\text{O}_4 \)
• \( \text{ZnS}_2\text{O}_4 + \text{Na}_2\text{CO}_3 \rightarrow \text{ZnCO}_3 + \text{Na}_2\text{S}_2\text{O}_4 \)

These describe the sequence of reactions used to produce sodium hydrosulfite.

• The first reaction shows how zinc and sulfur dioxide form a precursor compound, \( \text{ZnS}_2\text{O}_4 \).
• The second equation depicts producing the desired \( \text{Na}_2\text{S}_2\text{O}_4 \) from this precursor.

Balanced equations help in stoichiometry, which involves calculating the quantities of reactants and products. It is fundamental for determining things like limiting reactants and predicting yields.