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Chapter 9: Problem 7

Sodium chloride is produced from its elements through a synthesis reaction. What mass of each reactant would be required to produce 25.0 \(\mathrm{mol}\) of sodium chloride?

Short Answer

Expert verified
You need 575 g of sodium and 886.25 g of chlorine gas to produce 25.0 moles of sodium chloride.

Step by step solution

01

- Write the Balanced Chemical Equation

The first step is to write the balanced chemical equation for the synthesis of sodium chloride from its elements. The reaction is:\[\text{2 Na} (s) + \text{Cl}_2 (g) \rightarrow 2 \text{NaCl} (s)\]
02

- Determine Moles of Reactants Needed

The balanced equation shows that 2 moles of sodium (\text{Na}) react with 1 mole of chlorine gas (\text{Cl}_2) to produce 2 moles of sodium chloride (\text{NaCl}). To find the moles of each reactant required to produce 25.0 moles of \text{NaCl}, set up the stoichiometric relationships:For sodium: \[\text{moles of Na} = 25.0 \text{ mol NaCl} \times \frac{2 \text{ mol Na}}{2 \text{ mol NaCl}} = 25.0 \text{ mol Na}\]For chlorine gas: \[\text{moles of } \text{Cl}_2 = 25.0 \text{ mol NaCl} \times \frac{1 \text{ mol } \text{Cl}_2}{2 \text{ mol NaCl}} = 12.5 \text{ mol } \text{Cl}_2\]
03

- Convert Moles to Mass

Now convert the moles of each reactant to grams. The atomic mass of sodium (\text{Na}) is 23.0 \text{ g/mol} and the molecular mass of chlorine gas (\text{Cl}_2) is 70.9 \text{ g/mol}.For sodium: \[\text{mass of Na} = 25.0 \text{ mol Na} \times 23.0 \text{ g/mol} = 575 \text{ g Na}\]For chlorine gas: \[\text{mass of } \text{Cl}_2 = 12.5 \text{ mol } \text{Cl}_2 \times 70.9 \text{ g/mol} = 886.25 \text{ g } \text{Cl}_2\]

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Key Concepts

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

balanced chemical equation
A balanced chemical equation shows the reactants and products in a chemical reaction with their correct proportions. Ensuring the equation is balanced means the number of atoms for each element is the same on both the reactant and product sides. This is essential to adhere to the Law of Conservation of Mass, which states that matter cannot be created or destroyed in a chemical reaction. For the synthesis of sodium chloride:2 Na (s) + Cl_2 (g) → 2 NaCl (s)Notice we have 2 sodium (Na) atoms and 2 chlorine (Cl) atoms on each side. Balanced equations are crucial for correctly determining stoichiometry and moles of substances involved.
stoichiometry
Stoichiometry involves the calculation of reactants and products in chemical reactions. By using a balanced chemical equation, we can establish the relationships between the amounts of each substance, typically measured in moles. In our sodium chloride synthesis equation, 2 moles of Na react with 1 mole of Cl_2 to produce 2 moles of NaCl. Stoichiometry is crucial when we need to scale reactions up or down, or when we wish to convert between moles of different substances. It is essentially a way to predict the quantities of chemicals consumed and produced.
mole-to-mass conversion
The mole-to-mass conversion allows us to convert between the number of moles and the mass of a substance. Using the molar mass of the substance (grams per mole), we can calculate the mass required for a certain number of moles. For sodium (Na), the molar mass is 23.0 g/mol, and for chlorine gas (Cl_2), it's 70.9 g/mol. For example, if we need 25.0 moles of Na:mass of Na = 25.0 mol × 23.0 g/mol = 575 g Na. Similarly,mass of Cl_2 = 12.5 mol × 70.9 g/mol = 886.25 g Cl_2.Mole-to-mass conversion is key for practical applications like lab preparations.
reaction stoichiometry
Reaction stoichiometry involves using the balanced chemical equation to determine the relationships between reactants and products in a reaction. It's about understanding how much of each reactant is needed to produce a desired amount of product. For example, to produce 25.0 moles of NaCl:For Na: 25.0 mol NaCl × (2 mol Na / 2 mol NaCl) = 25.0 mol Na. For Cl_2: 25.0 mol NaCl × (1 mol Cl_2 / 2 mol NaCl) = 12.5 mol Cl_2. This tells us that to produce 25.0 moles of NaCl, we need 25.0 moles of Na and 12.5 moles of Cl_2. Understanding these stoichiometric relationships helps ensure that the reactions are efficient and that the correct amounts of reagents are used.

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Most popular questions from this chapter

a. Distinguish between the theoretical yield and actual yield in stoichiometric calculations. b. How does the value of the theoretical yield generally compare with the value of the actual yield?

As early as \(1938,\) the use of NaOH was suggested as a means of removing \(\mathrm{CO}_{2}\) from the cabin of a spacecraft according to the following (unbalanced) reaction: \(\mathrm{NaOH}+\mathrm{CO}_{2} \longrightarrow \mathrm{Na}_{2} \mathrm{CO}_{3}+\mathrm{H}_{2} \mathrm{O}\) . a. If the average human body discharges 925.0 \(\mathrm{gCO}_{2}\) per day, how many moles of NaOH are needed each day for each person in the spacecraft? b. How many moles of each product are formed?

The energy used to power one of the Apollo lunar missions was supplied by the following overall reaction: \(2 \mathrm{N}_{2} \mathrm{H}_{4}+\left(\mathrm{CH}_{3}\right)_{2} \mathrm{N}_{2} \mathrm{H}_{2}+3 \mathrm{N}_{2} \mathrm{O}_{4} \longrightarrow\) \(6 \mathrm{N}_{2}+2 \mathrm{CO}_{2}+8 \mathrm{H}_{2} \mathrm{O}\) . For the phase of the mission when the lunar module ascended from the surface of the moon, a total of \(1200 . \mathrm{kg} \mathrm{N}_{2} \mathrm{H}_{4}\) was available to react with \(1000 . \mathrm{kg}\left(\mathrm{CH}_{3}\right)_{2} \mathrm{N}_{2} \mathrm{H}_{2}\) and \(4500 . \mathrm{kg} \mathrm{N}_{2} \mathrm{O}_{4}\) a. For this portion of the flight, which of the allocated components was used up first? b. How much water, in kilograms, was put into the lunar atmosphere through this reaction?

Analyzing Results In the lab, you run an experiment that appears to have a percentage yield of 115%. Propose reasons for this result. Can an actual yield ever exceed a theoretical yield? Explain your answer.

a. Explain the concept of mole ratio as used in reaction stoichiometry problems. b. What is the source of this ratio?
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