Problem 56 - Chemistry and Chemical Reactivity

Question

Iodine is made by the reaction $$\begin{aligned} 2 \mathrm{NaIO}_{3}(\mathrm{aq})+5 \mathrm{NaHSO}_{3}(\mathrm{aq}) & \longrightarrow \\\3 \mathrm{NaHSO}_{4}(\mathrm{aq})+2 \mathrm{Na}_{2} \mathrm{SO}_{4}(\mathrm{aq}) &+\mathrm{H}_{2} \mathrm{O}(\ell)+\mathrm{I}_{2}(\mathrm{aq}) \end{aligned}$$ (a) Name the two reactants. (b) If you wish to prepare \(1.00 \mathrm{kg}\) of \(\mathrm{I}_{2},\) what mass of NalO \(_{3}\) is required? What mass of \(\mathrm{NaHSO}_{3} ?\)

Step-by-Step Solution

Answer

The reactants are sodium iodate and sodium bisulfite. 1558.37 g of NaIO the 3 is needed, and 2051.38 g of NaHSO the 3 is needed.

Step 1: Identify Reactants in the Equation

The chemical equation provided is: \[2 \mathrm{NaIO}_{3}(\mathrm{aq}) + 5 \mathrm{NaHSO}_{3}(\mathrm{aq}) \rightarrow 3 \mathrm{NaHSO}_{4}(\mathrm{aq}) + 2 \mathrm{Na}_{2} \mathrm{SO}_{4}(\mathrm{aq}) + \mathrm{H}_{2} \mathrm{O}(\ell) + \mathrm{I}_{2}(\mathrm{aq})\]. The reactants in this equation are \(\mathrm{NaIO}_{3}\) (sodium iodate) and \(\mathrm{NaHSO}_{3}\) (sodium bisulfite).

Step 2: Calculate Molar Mass of Iodine

Since we need to produce 1.00 kg of \(\mathrm{I}_{2}\), we first calculate its molar mass. The molar mass of iodine (\(\mathrm{I}_{2}\)) is calculated as follows: each iodine atom has a molar mass of \(126.90\) g/mol, so \(\mathrm{I}_{2}\) = \(2 \times 126.90 = 253.80\) g/mol.

Step 3: Convert Desired Mass of Iodine to Moles

Convert the mass of iodine to moles using its molar mass: \[1.00 \text{ kg of } \mathrm{I}_{2} = 1000 \text{ g}\]. Therefore, the number of moles is: \[\text{moles of I}_2 = \frac{1000 \text{ g}}{253.80 \text{ g/mol}} \approx 3.94 \text{ mol}\].

Step 4: Stoichiometry of Sodium Iodate

From the balanced equation, \(2\) moles of \(\mathrm{NaIO}_{3}\) produce \(1\) mole of \(\mathrm{I}_{2}\). So, to produce \(3.94\) moles of \(\mathrm{I}_{2}\), the moles of \(\mathrm{NaIO}_{3}\) required are: \[2 \times 3.94 = 7.88 \text{ moles of } \mathrm{NaIO}_{3}\].

Step 5: Calculate Mass of Sodium Iodate Required

The molar mass of \(\mathrm{NaIO}_{3}\) is calculated as: \(\mathrm{Na} = 22.99\), \(\mathrm{I} = 126.90\), \(3\times \mathrm{O} = 48.00\). Thus, \(\mathrm{NaIO}_{3}\) is \(22.99 + 126.90 + 48.00 = 197.89\) g/mol. Therefore, the mass needed is: \[7.88 \text{ moles} \times 197.89 \text{ g/mol} \approx 1558.37 \text{ g of } \mathrm{NaIO}_{3}\].

Step 6: Stoichiometry of Sodium Bisulfite

From the balanced equation, \(5\) moles of \(\mathrm{NaHSO}_{3}\) are needed to produce \(1\) mole of \(\mathrm{I}_{2}\). To produce \(3.94\) moles of \(\mathrm{I}_{2}\), the moles of \(\mathrm{NaHSO}_{3}\) required are: \[5 \times 3.94 = 19.7 \text{ moles of } \mathrm{NaHSO}_{3}\].

Step 7: Calculate Mass of Sodium Bisulfite Required

Calculate the molar mass of \(\mathrm{NaHSO}_{3}\): \(\mathrm{Na} = 22.99\), \(\mathrm{H} = 1.01\), \(\mathrm{S} = 32.07\), \(3 \times \mathrm{O} = 48.00\). Thus, \(\mathrm{NaHSO}_{3}\) is \(22.99 + 1.01 + 32.07 + 48.00 = 104.07\) g/mol. Therefore, the mass needed is: \[19.7 \text{ moles} \times 104.07 \text{ g/mol} \approx 2051.38 \text{ g of } \mathrm{NaHSO}_{3}\].

Key Concepts

Chemical Reactions

Chemical reactions involve the breaking and forming of chemical bonds, leading to the transformation of reactants into products. In our exercise, two reactants, sodium iodate \(\mathrm{NaIO}_3\) and sodium bisulfite \(\mathrm{NaHSO}_3\), undergo a chemical reaction to produce iodine \(\mathrm{I}_2\) along with other products.

• A chemical equation represents a chemical reaction, indicating the substances involved in the reaction and their stoichiometric relationships.
• The coefficients in a balanced equation tell us the ratio of moles of each reactant and product involved in the reaction. Here, 2 moles of \(\mathrm{NaIO}_3\) react with 5 moles of \(\mathrm{NaHSO}_3\) to produce 1 mole of \(\mathrm{I}_2\).
• Understanding stoichiometry helps predict how much of a reactant is needed to form a desired amount of product.

This balanced equation is crucial for calculating the reactants needed to produce a specific amount of iodine.

Molar Mass Calculation

Molar mass is an essential concept in stoichiometry, providing the link between the mass of substances and the numbers of moles involved. The molar mass is calculated by summing the masses of all atoms in a molecule, measured in grams per mole (g/mol).

• To calculate the molar mass of a compound, sum the molar masses of all the atoms in its molecular formula. For instance, the molar mass of iodine \(\mathrm{I}_2\) is calculated by adding the molar masses of two iodine atoms.
• Each iodine atom has a molar mass of 126.90 g/mol, so \(\mathrm{I}_2\) has a molar mass of \(2 \times 126.90 = 253.80\) g/mol.

This value is used to convert between the mass of a substance and the number of moles, which is crucial in stoichiometric calculations. A practical understanding of molar mass allows for precise ingredient measurement in chemical reactions.

Iodine Production

Iodine is an essential element in various industrial applications and chemical reactions. Producing iodine typically involves using compounds like sodium iodate, as demonstrated in the exercise.

• One important aspect of producing iodine is calculating the amount of reactant needed, which involves stoichiometry.
• In the given exercise, the goal is to produce \(1.00\) kg of \(\mathrm{I}_2\), prompting the need to determine how much \(\mathrm{NaIO}_3\) and \(\mathrm{NaHSO}_3\) are required to achieve this output.
• Applying stoichiometry, you convert the desired mass of iodine to moles, identify the stoichiometric ratios from the balanced equation, and calculate the mass of each required reactant.

Understanding this process is essential in optimizing reactions for iodine production, ensuring an efficient and effective synthesis.