/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 63 When an electric current is pass... [FREE SOLUTION] | 91影视

91影视

Log out

Learning Materials

Features

Discover

Chapter 4: Problem 63

When an electric current is passed through an aqueous solution of \(\mathrm{NaCl}\), the valuable industrial chemicals \(\mathrm{H}_{2}(\mathrm{g}), \mathrm{Cl}_{2}(\mathrm{g}),\) and \(\mathrm{NaOH}\) are produced. $$2 \mathrm{NaCl}(\mathrm{aq})+2 \mathrm{H}_{2} \mathrm{O}(\ell) \rightarrow \mathrm{H}_{2}(\mathrm{g})+\mathrm{Cl}_{2}(\mathrm{g})+2 \mathrm{NaOH}(\mathrm{aq})$$ What mass of NaOH can be formed from 15.0 L of 0.35 M NaCl? What mass of chlorine is obtained?

Short Answer

Expert verified
210 g of NaOH and 186.61 g of Cl2 are produced.

Step by step solution

01

Determine Moles of NaCl

First, we need to calculate the moles of NaCl in the solution. We use the formula for moles: \( ext{Moles} = ext{Molarity} \times ext{Volume} \). Since we have 0.35 M NaCl and a volume of 15.0 L, we find the moles as follows: \( ext{Moles of NaCl} = 0.35 \, M \times 15.0 \, L = 5.25 \, ext{moles} \).
02

Use Stoichiometry for NaOH

The balanced chemical equation shows that 2 moles of NaCl produce 2 moles of NaOH. Therefore, the moles of NaOH will be the same as the moles of NaCl, which is 5.25 moles.
03

Calculate Mass of NaOH

The formula to calculate mass is \( ext{Mass} = ext{Moles} \times ext{Molar Mass} \). The molar mass of NaOH is approximately 40.00 g/mol. Therefore, the mass of NaOH formed is \( 5.25 \, ext{moles} \times 40.00 \, ext{g/mol} = 210 \, ext{g} \).
04

Use Stoichiometry for Chlorine Gas

According to the balanced equation, 2 moles of NaCl produce 1 mole of \( \mathrm{Cl}_{2} \). This means from 5.25 moles of NaCl, we can produce \( \frac{5.25}{2} = 2.625 \, ext{moles of} \, \mathrm{Cl}_{2} \).
05

Calculate Mass of Chlorine Gas

The molar mass of \( \mathrm{Cl}_{2} \) is approximately 70.90 g/mol. Therefore, the mass of chlorine gas produced is \( 2.625 \, ext{moles} \times 70.90 \, ext{g/mol} = 186.61 \, ext{g} \).

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Over 30 million students worldwide already upgrade their learning with 91影视!

Key Concepts

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

Understanding Molarity
Molarity is a way to express the concentration of a solution. It's calculated as the number of moles of solute per liter of solution. For example, in the given problem, the solution has a molarity of 0.35 M, meaning there are 0.35 moles of sodium chloride (NaCl) per liter of solution.

To find out how many moles you have when given the volume of the solution, simply multiply the molarity by the solution's volume in liters. In this case, with 15 liters of 0.35 M NaCl, the calculation is:
  • Moles of NaCl = 0.35 M 脳 15 L = 5.25 moles
Understanding molarity helps you determine how much of a chemical is present in a given volume, which is crucial in reactions like electrolysis.
Basics of Stoichiometry
Stoichiometry is the study of the quantitative relationships in chemical reactions. It helps us figure out how much of each substance is consumed and produced during a reaction.

In the exercise, the balanced equation is:
  • 2 NaCl + 2 H\(_2\)O 鈫� H\(_2\) + Cl\(_2\) + 2 NaOH
This tells us:
  • 2 moles of NaCl will produce 2 moles of NaOH.
  • 2 moles of NaCl will produce 1 mole of Cl\(_2\).
Knowing these relationships allows us to calculate the amount of products formed. For instance, from 5.25 moles of NaCl:
  • You can make 5.25 moles of NaOH because of the 1:1 ratio.
  • You can produce 2.625 moles of Cl\(_2\) because of the 2:1 ratio.
Stoichiometry is like a recipe for chemistry, ensuring you know the right proportions of ingredients.
Interpreting Chemical Equations
Chemical equations provide a symbolic representation of chemical reactions. They're balanced to conserve mass and show the correct proportions of reactants and products.

The equation given in the problem:
  • 2 NaCl(aq) + 2 H\(_2\)O(l) 鈫� H\(_2\)(g) + Cl\(_2\)(g) + 2 NaOH(aq)
This tells us that:
  • 2 sodium chloride molecules are needed for the reaction to proceed.
  • Water is involved but isn't in the products as a separate entity.
  • The reactants result in the formation of hydrogen gas, chlorine gas, and sodium hydroxide.
Balancing equations ensures no atoms are lost or gained, maintaining the law of conservation of mass. This balance helps understand the moles and mass of each substance involved.

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

What mass of \(\mathrm{Na}_{2} \mathrm{CO}_{3},\) in grams, is required for complete reaction with \(50.0 \mathrm{mL}\) of \(0.125 \mathrm{M}\) \(\mathrm{HNO}_{3} ?\) $$\mathrm{Na}_{2} \mathrm{CO}_{3}(\mathrm{aq})+2 \mathrm{HNO}_{3}(\mathrm{aq}) \rightarrow 2 \mathrm{NaNO}_{3}(\mathrm{aq})+\mathrm{CO}_{2}(\mathrm{g})+\mathrm{H}_{2} \mathrm{O}(\ell)$$

A mixture of butene, \(\mathrm{C}_{4} \mathrm{H}_{8},\) and butane, \(\mathrm{C}_{4} \mathrm{H}_{10}\) is burned in air to give \(\mathrm{CO}_{2}\) and water. Suppose you burn 2.86 g of the mixture and obtain \(8.80 \mathrm{g}\) of \(\mathrm{CO}_{2}\) and \(4.14 \mathrm{g}\) of \(\mathrm{H}_{2} \mathrm{O}\). What are the mass percentages of butene and butane in the mixture?

ATOM ECONOMY: Ethylene oxide, \(\mathrm{C}_{2} \mathrm{H}_{4} \mathrm{O},\) is an important industrial chemical [as it is the starting place to make such important chemicals as ethylene glycol (antifreeze) and various polymers One way to make the compound is called the "chlorohydrin route." $$\mathrm{C}_{2} \mathrm{H}_{4}+\mathrm{Cl}_{2}+\mathrm{Ca}(\mathrm{OH})_{2} \rightarrow \mathrm{C}_{2} \mathrm{H}_{4} \mathrm{O}+\mathrm{CaCl}_{2}+\mathrm{H}_{2} \mathrm{O}$$ Another route is the modern catalytic reaction. $$\mathrm{C}_{2} \mathrm{H}_{4}+1 / 2 \mathrm{O}_{2} \rightarrow \mathrm{C}_{2} \mathrm{H}_{4} \mathrm{O}$$ (a) Calculate the \% atom economy for the production of \(\mathrm{C}_{2} \mathrm{H}_{4} \mathrm{O}\) in each of these reactions. Which is the more efficient method? (b) What is the percent yield of \(\mathrm{C}_{2} \mathrm{H}_{4} \mathrm{O}\) if 867 g of \(\mathrm{C}_{2} \mathrm{H}_{4}\) is used to synthesize \(762 \mathrm{g}\) of the product by the catalytic reaction?

Oyster beds in the oceans require chloride ions for growth. The minimum concentration is \(8 \mathrm{mg} / \mathrm{L}\) (8 parts per million). To analyze for the amount of chloride ion in a 50.0 -mL sample of water, you add a few drops of aqueous potassium chromate and then titrate the sample with \(25.60 \mathrm{mL}\) of 0.001036 M silver nitrate. The silver nitrate reacts with chloride ion, and, when the ion is completely removed, the silver nitrate reacts with potassium chromate to give a red precipitate. (a) Write a balanced net ionic equation for the reaction of silver nitrate with chloride ions. (b) Write a complete balanced equation and a net ionic equation for the reaction of silver nitrate with potassium chromate, indicating whether each compound is water-soluble or not. (c) What is the concentration of chloride ions in the sample? Is it sufficient to promote oyster growth?

A saturated solution of milk of magnesia, \(\mathrm{Mg}(\mathrm{OH})_{2},\) has a pH of \(10.5 .\) What is the hydronium ion concentration of the solution? Is the solution acidic or basic?
See all solutions

Recommended explanations on Chemistry Textbooks

Ionic and Molecular Compounds

Read Explanation

Nuclear Chemistry

Read Explanation

Inorganic Chemistry

Read Explanation

Physical Chemistry

Read Explanation

Chemical Analysis

Read Explanation

Kinetics

Read Explanation
View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.