Question

Iodine is made by the following reaction \(2 \mathrm{NaIO}_{3}(\mathrm{aq})+5 \mathrm{NaHSO}_{3}(\mathrm{aq}) \rightarrow\) $$ 3 \mathrm{NaHSO}_{4}(\mathrm{aq})+2 \mathrm{Na}_{2} \mathrm{SO}_{4}(\mathrm{aq})+\mathrm{H}_{2} \mathrm{O}(\ell)+\mathrm{I}_{2}(\mathrm{aq}) $$ (a) Name the two reactants. (b) If you wish to prepare \(1.00 \mathrm{kg}\) of \(\mathrm{I}_{2},\) what masses of \(\mathrm{NalO}_{3}\) and \(\mathrm{NaHSO}_{3}\) are required? (c) What is the theoretical yield of \(I_{2}\) if you mixed \(15.0 \mathrm{g}\) of \(\mathrm{NaIO}_{3}\) with \(125 \mathrm{mL}\) of \(0.853 \mathrm{M} \mathrm{NaHSO}_{3} ?\)

Short answer

(a) Sodium iodate and sodium bisulfite. (b) 1188 g NaIO3 and 2050 g NaHSO3. (c) 5.41 g I2.

Step-by-step solution

Identify Reactants

The equation given is: \[2 \text{NaIO}_3(\text{aq}) + 5 \text{NaHSO}_3(\text{aq}) \rightarrow 3 \text{NaHSO}_4(\text{aq}) + 2 \text{Na}_2\text{SO}_4(\text{aq}) + \text{H}_2\text{O}(\ell) + \text{I}_2(\text{aq})\]The two reactants in this chemical reaction are sodium iodate (\(\text{NaIO}_3\)) and sodium bisulfite (\(\text{NaHSO}_3\)).

Calculate Molar Mass of Iodine

The molar mass of iodine \((\text{I}_2)\) is calculated as follows:\[2 \times \text{Atomic mass of I} = 2 \times 126.90 = 253.80\, \text{g/mol}\]We need 1.00 kg of \(\text{I}_2\), which is equivalent to 1000 g.

Determine Moles of Iodine Needed

Calculate the moles of \(\text{I}_2\) needed:\[\text{Moles of I}_2 = \frac{1000\, \text{g}}{253.80\, \text{g/mol}} \approx 3.94\, \text{mol}\]

Use Stoichiometry for Reactants

According to the balanced equation, for every 2 moles of \(\text{NaIO}_3\) and 5 moles of \(\text{NaHSO}_3\), 1 mole of \(\text{I}_2\) is produced.Thus, we need:- \(2 \times 3.94 = 7.88\) moles of \(\text{NaIO}_3\)- \(5 \times 3.94 = 19.7\) moles of \(\text{NaHSO}_3\)

Calculate Masses of NaIO3 and NaHSO3

Calculate the molar masses:- \(\text{NaIO}_3 = 150.89\, \text{g/mol}\)- \(\text{NaHSO}_3 = 104.06\, \text{g/mol}\)Then, calculate the masses:- Mass of \(\text{NaIO}_3 = 7.88\, \text{mol} \times 150.89\, \text{g/mol} \approx 1188\, \text{g}\)- Mass of \(\text{NaHSO}_3 = 19.7\, \text{mol} \times 104.06\, \text{g/mol} \approx 2050\, \text{g}\)

Calculate Moles of NaHSO3 in Solution

Calculate the moles of \(\text{NaHSO}_3\) in the solution:\[\text{Moles of NaHSO}_3 = (0.853\, \text{mol/L}) \times (0.125\, \text{L}) = 0.106625\, \text{mol}\]

Determine Limiting Reactant

Based on the reaction stoichiometry:- \(\text{NaIO}_3\) available: \(\frac{15.0\, \text{g}}{150.89\, \text{g/mol}} \approx 0.0994\, \text{mol}\)- \(\text{NaHSO}_3\) required to react with \(0.0994\, \text{mol}\) \(\text{NaIO}_3\): \(0.0994 \times \frac{5}{2} = 0.2485\, \text{mol}\)Given only \(0.106625\, \text{mol}\) \(\text{NaHSO}_3\), \(\text{NaHSO}_3\) is the limiting reactant.

Calculate Theoretical Yield of Iodine

Determine the moles of \(\text{I}_2\) producible by \(\text{NaHSO}_3\):\[\text{Moles of I}_2 = \frac{0.106625}{5} \approx 0.021325\, \text{mol}\]Hence, the theoretical mass of \(\text{I}_2\) is:\[\text{Mass of I}_2 = 0.021325\, \text{mol} \times 253.80\, \text{g/mol} \approx 5.41\, \text{g}\]

Key concepts

Limiting Reactant

In any chemical reaction, the limiting reactant is the substance that is entirely used up first. This reactant determines the maximum amount of product that can be formed. In the reaction of sodium iodate (NaIO extsubscript{3}) and sodium bisulfite (NaHSO extsubscript{3}), there is a competition to see which one gets used up first, limiting the reaction's progress.

The concept stems from the stoichiometry of the chemical equation. Here, for every 2 moles of NaIO extsubscript{3} and 5 moles of NaHSO extsubscript{3}, 1 mole of iodine ( ext{I} extsubscript{2}) is produced. Therefore, if we start with less NaHSO extsubscript{3} than needed to react with the available NaIO extsubscript{3}, NaHSO extsubscript{3} becomes the limiting reactant.

To find out which one is limiting, calculate the moles of each reactant available. For NaHSO extsubscript{3}, if we have fewer moles than required by the mole ratio, NaHSO extsubscript{3} will limit the amount of ext{I} extsubscript{2} produced. In practice, identifying the limiting reactant helps predict the theoretical yield of the desired product.

Theoretical Yield

The theoretical yield is the amount of product expected from a chemical reaction if everything goes perfectly, with no losses or side reactions. It represents the maximum amount of product that can be generated as calculated based on the limiting reactant.

Using the reaction of NaIO extsubscript{3} and NaHSO extsubscript{3}, once we identify the limiting reactant (NaHSO extsubscript{3} in this case), we know it will dictate the total ext{I} extsubscript{2} produced. Theoretical yield is strictly calculated by using stoichiometry based on quantities of the limiting reactant and the balanced chemical equation.

For instance, with 0.106625 moles of NaHSO extsubscript{3}, the theoretical moles of ext{I} extsubscript{2} are found by dividing by the mole ratio, giving us 0.021325 moles of ext{I} extsubscript{2}. To get the mass, multiply by the molar mass of ext{I} extsubscript{2}. Thus, the calculated theoretical yield is essential in laboratory settings as it provides a target for how much product to expect.

Molar Mass

Understanding molar mass is fundamental to performing stoichiometric calculations correctly. The molar mass of a compound is the mass of one mole of that substance and is expressed in grams per mole ( ext{g/mol}). It is determined by summing the atomic masses of all the atoms in a molecule.

For iodine ( ext{I} extsubscript{2}), you calculate its molar mass as follows:

• Atomic mass of iodine = 126.90 ext{g/mol}
• Molar mass of ext{I} extsubscript{2} = 2 × 126.90 = 253.80 ext{g/mol}

The molar mass allows you to convert grams to moles, which is crucial for determining amounts in reactions.

When dealing with equations, converting between masses and moles using molar mass helps track how much of each reactant is needed and how much product can realistically be produced. It bridges the world of chemical equations with real-world mass measurements, ensuring precise calculations of theoretical yields and identifying limiting reactants.