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Chapter 11: Problem 32

Iodine is prepared both in the laboratory and commercially by adding \(\mathrm{Cl}_{2}(g)\) to an agueous solution containing sodium iodide according to $$ 2 \mathrm{NaI}(a q)+\mathrm{Cl}_{2}(g) \rightarrow \mathrm{I}_{2}(s)+2 \mathrm{NaCl}(a q) $$ How many grams of sodium iodide must be used to produce \(50.0\) grams of iodine?

Short Answer

Expert verified
Approximately 59.06 grams of sodium iodide are needed.

Step by step solution

01

Write down the chemical reaction

The balanced chemical equation is given as:\[ 2 \text{NaI}(aq) + \text{Cl}_{2}(g) \rightarrow \text{I}_{2}(s) + 2 \text{NaCl}(aq) \]
02

Find the molar mass of iodine

The molar mass of iodine \( \text{I}_{2} \) is calculated using the atomic mass of iodine \( \approx 126.90 \) g/mol:\[ \text{Molar mass of } \text{I}_{2} = 2 \times 126.90 = 253.80 \text{ g/mol} \]
03

Calculate moles of iodine

To find the moles of iodine produced, use its mass in grams:\[ \text{Moles of } \text{I}_{2} = \frac{50.0 \text{ g}}{253.80 \text{ g/mol}} \approx 0.197 \text{ mol} \]
04

Use stoichiometry to find moles of sodium iodide

According to the balanced chemical equation, 2 moles of \( \text{NaI} \) produce 1 mole of \( \text{I}_{2} \). Therefore, the moles of \( \text{NaI} \) needed are double the moles of \( \text{I}_{2} \):\[ \text{Moles of } \text{NaI} = 2 \times 0.197 \text{ mol} = 0.394 \text{ mol} \]
05

Find the molar mass of sodium iodide

The molar mass of sodium iodide \( \text{NaI} \) is calculated using the atomic masses of sodium (\( \approx 22.99 \) g/mol) and iodine (\( \approx 126.90 \) g/mol):\[ \text{Molar mass of } \text{NaI} = 22.99 + 126.90 = 149.89 \text{ g/mol} \]
06

Calculate grams of sodium iodide required

Finally, convert the moles of \( \text{NaI} \) to mass:\[ \text{Mass of } \text{NaI} = 0.394 \text{ mol} \times 149.89 \text{ g/mol} \approx 59.06 \text{ g} \]

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Chemical Reactions
Chemical reactions are processes in which substances, known as reactants, transform into new substances, called products. In the laboratory preparation of iodine, chlorine gas (\( \mathrm{Cl}_{2}(g) \)) reacts with sodium iodide solution (\( \mathrm{NaI}(aq) \)). The chemical equation for this reaction is already balanced:
  • \( 2 \mathrm{NaI}(aq) + \mathrm{Cl}_{2}(g) \rightarrow \mathrm{I}_{2}(s) + 2 \mathrm{NaCl}(aq) \)
A balanced equation is crucial because it ensures the conservation of mass and moles, highlighting the stoichiometry of the reaction.
This stoichiometry tells us the relationship between the numbers of moles of each reactant and product involved. For example, 2 moles of sodium iodide produce 1 mole of iodine. This stoichiometric ratio helps in understanding the quantities of reactants needed to form a desired amount of product.
Molar Mass
Molar mass is an essential concept in stoichiometry. It helps us convert between the mass of a substance and the moles of that substance. Molar mass is the mass of one mole of a substance, usually expressed in grams per mole. For example, the molar mass of diatomic iodine (\( \mathrm{I}_{2} \)) is calculated by adding the atomic masses of the iodine atoms:
  • Atomic mass of iodine: approximately 126.90 g/mol
  • Molar mass of \( \mathrm{I}_{2} \) = 2 x 126.90 = 253.80 g/mol
Understanding molar masses allows us to understand how much of one chemical is equivalent to one mole, which is critical when calculating how much of a reactant is needed or product is formed in a reaction.Similarly, calculating the molar mass of sodium iodide (\( \mathrm{NaI} \)) involves adding the atomic masses of sodium and iodine, which results in a molar mass of approximately 149.89 g/mol.
Mole Calculations
Mole calculations are a key part of stoichiometry and involve converting between mass, moles, and (sometimes) number of molecules. These calculations are grounded in the concept of the mole鈥攁 central unit in chemistry that links the macroscopic and atomic scales.For instance, when you have 50.0 grams of iodine, you can find how many moles of iodine are present by using its molar mass:
  • Moles of \( \mathrm{I}_{2} \) = \( \frac{50.0 \text{ g}}{253.80 \text{ g/mol}} \approx 0.197 \text{ mol} \)
Once you know the moles of iodine, you can use their stoichiometry to find the moles of sodium iodide needed. Since the reaction requires 2 moles of sodium iodide to produce 1 mole of iodine, you can determine the moles of sodium iodide:
  • Moles of \( \mathrm{NaI} \) = 2 x 0.197 mol = 0.394 mol
Finally, to find out how much sodium iodide mass is required, multiply the moles of sodium iodide by its molar mass:
  • Mass of \( \mathrm{NaI} \) = 0.394 mol x 149.89 g/mol \( \approx 59.06 \text{ g} \)
These calculations show the practical application of stoichiometry in predicting the amounts of substances needed or generated in chemical reactions.

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Most popular questions from this chapter

A mixture of \(\mathrm{Na}_{2} \mathrm{SO}_{4}(s)\) and \(\mathrm{NaHSO}_{4}(s)\) haviny a mass of \(2.606\) giams is dissulved in watet anul excess \(\mathrm{Ba}(\mathrm{OH})_{2}(s)\) is added, precipitating \(\mathrm{BaSO}_{4}(s)\) according to $$ \begin{aligned} \mathrm{Na}_{2} \mathrm{SO}_{4}(s)+\mathrm{Ba}(\mathrm{OH})_{2}(a q) \rightarrow & \mathrm{BaSO}_{4}(s)+2 \mathrm{NaOH}(a q) \\ \mathrm{NaHSO}_{4}(s)+\mathrm{Ba}(\mathrm{OH})_{2}(a q) \rightarrow \\ \mathrm{BaSO}_{4}(s)+\mathrm{NaOH}(a q)+\mathrm{H}_{2} \mathrm{O}(l) \end{aligned} $$ The precipitate, \(\mathrm{BaSO}_{4}(s)\), has a mass of \(4.688\) grams. Calculate the mass percentages of \(\mathrm{Na}_{2} \mathrm{SO}_{4}(s)\) and \(\mathrm{NaHSO}_{4}(s)\) in the mixture.

A \(3.78\) -gram sample of iron metal is reacted with sulfur to produce \(5.95\) grams of iron sulfide. Determine the empirical formula of this compound.

Naproxen sodium, \(\mathrm{C}_{14} \mathrm{H}_{13} \mathrm{O}_{3} \mathrm{Na}(s)\), the active ingredient in the pain reliever Aleve \(^{\circledast}\), is a nonsteroidal anti-inflammatory drug. Calculate the mass percentage of each of the elements in naproxen sodium to four significant figures and show that the sum of these mass percentages totals one hundred percent.

When left exposed to air and water, iron metal will rust, forming an oxide. If a bar of pure iron is allowed to rust and then both the rust and remaining iron are weighed, will the total mass be greater than, less than, or equal to the mass of the original iron bar? If all the rust is scraped off, will the mass of the bar now be greater than, less than, or equal to the original mass of the bar? Justify your answers.

The concept of determining which reactant is limiting and which is in excess is akin to determining the number of sandwiches that can be made from a set number of ingredients. Assuming that a cheese sandwich consists of 2 slices of bread and 3 slices of cheese, determine the number of cheese sandwiches that can be prepared from a loaf of 24 slices of bread and a package of 40 slices of cheese. Which of the two ingredients limits the number of sandwiches that can be made? What quantity of the ingredient in excess remains?
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