/*! 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 94 A chemical plant uses electrical... [FREE SOLUTION] | 91影视

91影视

Log out

Learning Materials

Features

Discover

Chapter 3: Problem 94

A chemical plant uses electrical energy to decompose aqueous solutions of \(\mathrm{NaCl}\) to give \(\mathrm{Cl}_{2}, \mathrm{H}_{2}\), and \(\mathrm{NaOH}\) : \(2 \mathrm{NaCl}(a q)+2 \mathrm{H}_{2} \mathrm{O}(l) \longrightarrow \underset{2 \mathrm{NaOH}(a q)}{\longrightarrow}+\mathrm{H}_{2}(g)+\mathrm{Cl}_{2}(g)\) If the plant produces \(1.5 \times 10^{6} \mathrm{~kg}\left(1500\right.\) metric tons) of \(\mathrm{Cl}_{2}\) daily, estimate the quantities of \(\mathrm{H}_{2}\) and \(\mathrm{NaOH}\) produced.

Short Answer

Expert verified
The estimated daily production of H鈧� is approximately \(4.26299 \times 10^4\ kg\) (42.63 metric tons), and the estimated daily production of NaOH is approximately \(1.6925 \times 10^6\ kg\) (1692.5 metric tons).

Step by step solution

01

Calculate the moles of Cl鈧� produced

Since we are given the mass of Cl鈧� produced daily, we can calculate the moles of Cl鈧� by dividing the mass by the molar mass of Cl鈧�. The molar mass of Cl鈧� is approximately 70.9 g/mol. First, convert the daily production of Cl鈧� to grams: \(1.5 \times 10^{6}\ kg \times \frac{1000\ g}{1\ kg} = 1.5 \times 10^9\ g\) Next, calculate the moles of Cl鈧�: \(\frac{1.5 \times 10^9\ g}{70.9\ g/mol} = 2.11563 \times 10^7\ moles\)
02

Determine the moles of H鈧� and NaOH produced

Now, use the stoichiometry of the balanced equation to determine the moles of H鈧� and NaOH produced. From the balanced equation, we can see that for every 1 mole of Cl鈧� produced, 1 mole of H鈧� and 2 moles of NaOH are produced: \(2 \mathrm{NaCl}(a q)+2 \mathrm{H}_{2} \mathrm{O}(l) \longrightarrow \underset{2 \mathrm{NaOH}(a q)}{\longrightarrow}+\mathrm{H}_{2}(g)+\mathrm{Cl}_{2}(g)\) So, the moles of H鈧� produced will be equal to the moles of Cl鈧�: \(2.11563 \times 10^7\ moles\ H鈧俓) And the moles of NaOH produced will be twice the moles of Cl鈧�: \(2 \times 2.11563 \times 10^7\ moles\ NaOH \approx 4.23126 \times 10^7\ moles\)
03

Calculate the mass of H鈧� and NaOH produced

Finally, we'll use the molar mass of H鈧� and NaOH to convert the moles calculated in Step 2 to mass: Molar mass of H鈧� = 2.016 g/mol Molar mass of NaOH = 40 g/mol Calculate the mass of H鈧� produced: \((2.11563 \times 10^7\ moles\ H鈧�) \times \frac{2.016\ g}{1\ mole\ H鈧倉 \approx 4.26299 \times 10^7\ g\) Convert to kg: \(\frac{4.26299 \times 10^7\ g\ H鈧倉{1000\ g/kg} \approx 4.26299 \times 10^4\ kg\ H鈧俓) Calculate the mass of NaOH produced: \((4.23126 \times 10^7\ moles\ NaOH) \times \frac{40\ g}{1\ mole\ NaOH} \approx 1.6925 \times 10^9\ g\ NaOH\) Convert to kg: \(\frac{1.6925 \times 10^9\ g\ NaOH}{1000\ g/kg} \approx 1.6925 \times 10^6\ kg\ NaOH\) So, the estimated daily production of H鈧� and NaOH is: H鈧�: \(4.26299 \times 10^4\ kg\) (approximately 42.63 metric tons) NaOH: \(1.6925 \times 10^6\ kg\) (approximately 1692.5 metric tons)

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.

Mole Concept
The mole concept is a fundamental principle in chemistry that functions as the bridge between the atomic world and the macroscopic world we experience. It allows chemists to count particles by weighing. In essence, a mole represents a fixed number of particles鈥攕pecifically, Avogadro's number, which is approximately 6.022 x 1023 particles.

Think of it like a 'chemist's dozen,' similar to how a dozen represents 12 of something, a mole represents 6.022 x 1023 particles. This is incredibly useful, as it means that if we know the molar mass of a substance (the mass of one mole of particles), we can easily convert between the mass of a substance and the number of particles or moles it represents. In the chemical reaction from our exercise, knowing the mole concept helps us calculate how much raw material is needed or how much product can be obtained.
Molar Mass Calculation
Molar mass calculation is simply determining the mass of one mole of a substance. The unit for molar mass is grams per mole (g/mol). It can be calculated by summing the atomic masses of all the atoms in a molecule. For instance, water (H2O) has a molar mass of approximately 18.015 g/mol because it has two hydrogen atoms (each with a molar mass of about 1.008 g/mol) and one oxygen atom (with a molar mass of about 15.999 g/mol).

In the example, the molar mass of Cl2 was used to convert the mass of chlorine gas produced into moles, a necessary step to then use our stoichiometric ratios. Accurate molar mass calculations are vital to stoichiometry since they ensure that the math regarding the number of particles involved in a reaction is correct.
Stoichiometric Calculations
Stoichiometric calculations involve using the balanced chemical equation to understand the proportions of reactants to products. It's all about the ratios in which substances react or form products. The balanced equation tells us the mole ratio, in this case, that for every 2 moles of NaCl and 2 moles of H2O, we get 2 moles of NaOH, 1 mole of H2, and 1 mole of Cl2.

With stoichiometry, once we know the amount of one substance in a reaction, we can calculate the amounts of all the other substances involved in the reaction. For the chemical plant problem, by knowing the daily production of Cl2, we estimated the quantities of H2 and NaOH produced. This is crucial, not just for assignments, but in real-world applications where chemists need to predict how much of each chemical is needed or will be produced in a reaction.

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

Hydrogen cyanide, \(\mathrm{HCN}\), is a poisonous gas. The lethal dose is approximately \(300 \mathrm{mg} \mathrm{HCN}\) per kilogram of air when inhaled. (a) Calculate the amount of HCN that gives the lethal dose in a small laboratory room measuring \(12 \times 15 \times 80 \mathrm{ft}\). The density of air at \(26{ }^{\circ} \mathrm{C}\) is \(0.00118 \mathrm{~g} / \mathrm{cm}^{3}\), (b) If the \(\mathrm{HCN}\) is formed by reaction of \(\mathrm{NaCN}\) with an acid such as \(\mathrm{H}_{2} \mathrm{SO}_{4}\), what mass of \(\mathrm{NaCN}\) gives the lethal dose in the room? \(2 \mathrm{NaCN}(s)+\mathrm{H}_{2} \mathrm{SO}_{4}(a q) \longrightarrow \mathrm{Na}_{2} \mathrm{SO}_{4}(a q)+2 \mathrm{HCN}(g)\) (c) HCN forms when synthetic fibers containing Orlon \({ }^{8}\) or Acrilan burn. Acrilan has an empirical formula of \(\mathrm{CH}_{2} \mathrm{CHCN}\), so \(\mathrm{HCN}\) is \(50.9 \%\) of the formula by mass. \(\mathrm{A}\) rug measures \(12 \times 15 \mathrm{ft}\) and contains \(30 \mathrm{oz}\) of Acrilan \(^{8}\) fibers per square yard of carpet. If the rug burns, will a lethal dose of \(\mathrm{HCN}\) be generated in the room? Assume that the yield of \(\mathrm{HCN}\) from the fibers is \(20 \%\) and that the carpet is \(50 \%\) consumed.

(a) What is the mass, in grams, of a mole of \({ }^{12} \mathrm{C}\) ? (b) How many carbon atoms are present in a mole of \({ }^{12} \mathrm{C}\) ?

Determine the formula weights of each of the following compounds: (a) nitrous oxide, \(\mathrm{N}_{2} \mathrm{O}\), known as laughing gas and used as an anesthetic in dentistry; (b) benzoic acid, \(\mathrm{HC}_{7} \mathrm{H}_{5} \mathrm{O}_{2}\), a substance used as a food preservative; (c) \(\mathrm{Mg}(\mathrm{OH})_{2}\), the active ingredient in milk of magnesia; (d) urea, \(\left(\mathrm{NH}_{2}\right)_{2} \mathrm{CO}\), a compound used as a nitrogen fertilizer; (e) isopentyl acetate, \(\mathrm{CH}_{3} \mathrm{CO}_{2} \mathrm{C}_{5} \mathrm{H}_{11}\), responsible for the odor of bananas.

Hydrogen sulfide is an impurity in natural gas that must be removed. One common removal method is called the Claus process, which relies on the reaction: $$ 8 \mathrm{H}_{2} \mathrm{~S}(g)+4 \mathrm{O}_{2}(g) \longrightarrow \mathrm{S}_{8}(l)+8 \mathrm{H}_{2} \mathrm{O}(g) $$ Under optimal conditions the Claus process gives \(98 \%\) yield of \(\mathrm{S}_{8}\) from \(\mathrm{H}_{2} \mathrm{~S}\). If you started with 300 grams of \(\mathrm{H}_{2} \mathrm{~S}\) and \(50.0\) grams of \(\mathrm{O}_{2}\), how many grams of \(\mathrm{S}_{\mathrm{g}}\) would be produced, assuming \(98 \%\) yield?

Calculate the following quantities: (a) mass, in grams, of \(0.105\) moles sucrose \(\left(\mathrm{C}_{12} \mathrm{H}_{22} \mathrm{O}_{11}\right)\) (b) moles of \(\mathrm{Zn}\left(\mathrm{NO}_{3}\right)_{2}\) in \(14350 \mathrm{~g}\) of this substance (c) number of molecules in \(1.0 \times 10^{-6} \mathrm{~mol} \mathrm{CH}_{3} \mathrm{CH}_{2} \mathrm{OH}\) (d) number of \(\mathrm{N}\) atoms in \(0.410 \mathrm{~mol} \mathrm{NH}_{3}\)
See all solutions

Recommended explanations on Chemistry Textbooks

Organic Chemistry

Read Explanation

Chemistry Branches

Read Explanation

Chemical Reactions

Read Explanation

Physical Chemistry

Read Explanation

Ionic and Molecular Compounds

Read Explanation

Inorganic Chemistry

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.