/*! 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 100 A chemical plant uses electrical... [FREE SOLUTION] | 91Ó°ÊÓ

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

Short Answer

Expert verified
The plant produces 42.7 metric tons of \(\mathrm{H}_2\) and 1692 metric tons of \(\mathrm{NaOH}\) daily.

Step by step solution

01

Calculate Moles of Chlorine Produced

First, we need to calculate the number of moles of chlorine gas, \(\mathrm{Cl}_{2}\), produced. The molar mass of \(\mathrm{Cl}_2\) is \(70.9 \, \mathrm{g/mol}\). So, the number of moles of \(\mathrm{Cl}_2\) produced is calculated by dividing the mass by the molar mass: \[ \text{Moles of } \mathrm{Cl}_{2} = \frac{1.5 \times 10^6 \text{ kg} \times 10^3 \text{ g/kg}}{70.9 \, \mathrm{g/mol}} = 2.115 \times 10^7 \, \mathrm{mol}. \]
02

Use Stoichiometry to Find Moles of Products

Refer to the stoichiometric coefficients from the balanced chemical equation: \(2 \mathrm{NaCl} + 2 \mathrm{H}_{2} \mathrm{O} \rightarrow 2 \mathrm{NaOH} + \mathrm{H}_{2} + \mathrm{Cl}_{2} \). For every mole of \(\mathrm{Cl}_{2}\) produced, the reaction produces 1 mole of \(\mathrm{H}_{2}\) and 2 moles of \(\mathrm{NaOH}\). Thus: \[ \text{Moles of } \mathrm{H}_{2} = 2.115 \times 10^7 \, \mathrm{mol}, \] \[ \text{Moles of } \mathrm{NaOH} = 2 \times 2.115 \times 10^7 = 4.230 \times 10^7 \, \mathrm{mol}. \]
03

Convert Moles to Mass

Finally, convert the moles of \(\mathrm{H}_{2}\) and \(\mathrm{NaOH}\) to mass. The molar mass of \(\mathrm{H}_{2}\) is \(2.02 \, \mathrm{g/mol}\) and \(\mathrm{NaOH}\) is \(40.00 \, \mathrm{g/mol}\). Calculate the mass: \[ \text{Mass of } \mathrm{H}_{2} = 2.115 \times 10^7 \, \mathrm{mol} \times 2.02 \, \mathrm{g/mol} = 4.27 \times 10^7 \, \mathrm{g} = 42.7 \, \text{metric tons}, \] \[ \text{Mass of } \mathrm{NaOH} = 4.230 \times 10^7 \, \mathrm{mol} \times 40.00 \, \mathrm{g/mol} = 1.692 \times 10^9 \, \mathrm{g} = 1692 \text{ metric tons}. \]

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Key Concepts

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

Electrolysis of NaCl
Electrolysis involves breaking down a compound using electrical energy. In the case of \(\mathrm{NaCl}\), electricity is used to decompose an aqueous solution into chlorine gas (\(\mathrm{Cl}_2\)), hydrogen gas (\(\mathrm{H}_2\)), and sodium hydroxide (\(\mathrm{NaOH}\)).
This process occurs in an electrolytic cell where two electrodes, an anode and a cathode, are immersed in the solution.
The \(\mathrm{NaCl}\) solution is split into its ions. At the cathode, water is reduced to produce hydrogen gas, while at the anode, chloride ions are oxidized to produce chlorine gas.
This results in the generation of hydroxide ions that combine with sodium ions to form sodium hydroxide. Electrolysis is widely used in industries for obtaining pure elements and compounds.
  • Cell Reaction: Decomposition takes place through electric current.
  • Products: \(\mathrm{H}_2\), \(\mathrm{Cl}_2\), and \(\mathrm{NaOH}\) as per the balanced reaction.
  • Applications: Used to produce alkali and chlorine products for various purposes.
Mole Calculations
Understanding mole calculations is essential for converting mass to moles, which is a fundamental step in stoichiometry.
The mole concept relates the amount of material in a sample to its mass and is a key concept in chemistry.
To calculate the number of moles of a substance, you divide the mass of the substance by its molar mass.
For example, determining moles of \(\mathrm{Cl}_2\) involves dividing the given mass by its molar mass of 70.9 \(\mathrm{g/mol}\). This gives you a sense of how many particles or molecules you are dealing with.
  • Step 1: Identify the substance and its molar mass.
  • Step 2: Divide the mass by the molar mass to find moles.
  • Considerations: Moles help relate the mass to the number of molecules or atoms in reactions.
Stoichiometric Coefficients
In a chemical equation, stoichiometric coefficients define the quantitative relationships between reactants and products.
They allow chemists to make precise calculations concerning the amounts of substances needed or produced in a reaction.
In the decomposition of \(\mathrm{NaCl}\), the stoichiometric coefficients are 2 for \(\mathrm{NaCl}\) and \(\mathrm{H}_2\mathrm{O}\), 1 for \(\mathrm{H}_2\) and \(\mathrm{Cl}_2\), and 2 for \(\mathrm{NaOH}\).
They indicate that two moles of \(\mathrm{NaCl}\) will yield one mole of \(\mathrm{Cl}_2\), one mole of \(\mathrm{H}_2\), and two moles of \(\mathrm{NaOH}\).
  • Essential for converting moles of substances in reactions.
  • Determines proportional relationships in chemical reactions.
  • Helps predict the amount of products generated from given reactants.
Mass-Mole Conversions
Converting between mass and moles is a common task when working with chemical equations.
It links the mass of substances to the number of moles through their molar masses.
The conversion process usually involves using the molar mass as a conversion factor.
Mass-mole calculations are crucial in determining how much of each reactant you need to start a reaction and in calculating expected product yields.
For instance, once the moles of \(\mathrm{H}_2\) or \(\mathrm{NaOH}\) are known, converting back to mass involves multiplying by the respective molar masses of \(2.02 \, \mathrm{g/mol}\) for \(\mathrm{H}_2\) and \(40.00 \, \mathrm{g/mol}\) for \(\mathrm{NaOH}\).
  • Mass to moles: Mass divided by molar mass gives moles.
  • Moles to mass: Moles multiplied by molar mass gives mass.
  • Application: Critical in manufacturing and laboratory settings to predict product output.

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Most popular questions from this chapter

Without doing any detailed calculations (but using a periodic table to give atomic weights), rank the following samples in order of increasing numbers of atoms: \(0.5 \mathrm{~mol} \mathrm{BCl}_{3}\) molecules, \(197 \mathrm{~g}\) gold, \(6.0 \times 10^{23} \mathrm{CCl}_{4}\) molecules.

Hydrofluoric acid, HF \((a q)\), cannot be stored in glass bottles because compounds called silicates in the glass are attacked by the \(\mathrm{HF}(a q)\). Sodium silicate \(\left(\mathrm{Na}_{2} \mathrm{SiO}_{3}\right)\), for example, reacts as follows: $$ \mathrm{Na}_{2} \mathrm{SiO}_{3}(s)+8 \mathrm{HF}(a q) \longrightarrow \mathrm{H}_{2} \mathrm{SiF}_{6}(a q)+2 \mathrm{NaF}(a q)+3 \mathrm{H}_{2} \mathrm{O}(l) $$ (a) How many moles of HF are needed to react with 0.300 mol of \(\mathrm{Na}_{2} \mathrm{SiO}_{3} ?\) (b) How many grams of NaF form when \(0.500 \mathrm{~mol}\) of HF reacts with excess \(\mathrm{Na}_{2} \mathrm{SiO}_{3} ?\) (c) How many grams of \(\mathrm{Na}_{2} \mathrm{SiO}_{3}\) can react with \(0.800 \mathrm{~g}\) of HF?

A mixture containing \(\mathrm{KClO}_{3}, \mathrm{~K}_{2} \mathrm{CO}_{3}, \mathrm{KHCO}_{3},\) and \(\mathrm{KCl}\) was heated, producing \(\mathrm{CO}_{2}, \mathrm{O}_{2},\) and \(\mathrm{H}_{2} \mathrm{O}\) gases according to the following equations: $$ \begin{aligned} 2 \mathrm{KClO}_{3}(s) & \longrightarrow 2 \mathrm{KCl}(s)+3 \mathrm{O}_{2}(g) \\\ 2 \mathrm{KHCO}_{3}(s) & \longrightarrow \mathrm{K}_{2} \mathrm{O}(s)+\mathrm{H}_{2} \mathrm{O}(g)+2 \mathrm{CO}_{2}(g) \\ \mathrm{K}_{2} \mathrm{CO}_{3}(s) & \longrightarrow \mathrm{K}_{2} \mathrm{O}(s)+\mathrm{CO}_{2}(g) \end{aligned} $$ The KCl does not react under the conditions of the reaction. If \(100.0 \mathrm{~g}\) of the mixture produces \(1.80 \mathrm{~g}\) of \(\mathrm{H}_{2} \mathrm{O}, 13.20 \mathrm{~g}\) of \(\mathrm{CO}_{2}\), and \(4.00 \mathrm{~g}\) of \(\mathrm{O}_{2},\) what was the composition of the original mixture? (Assume complete decomposition of the mixture.)

Determine the empirical formula of each of the following compounds if a sample contains (a) \(3.92 \mathrm{~mol} \mathrm{C}, 5.99 \mathrm{~mol} \mathrm{H}\) and \(2.94 \mathrm{~mol} \mathrm{O} ;(\mathbf{b}) 12.0 \mathrm{~g}\) calcium and \(2.8 \mathrm{~g}\) nitrogen; \((\mathbf{c})\) \(89.14 \%\) Au and \(10.86 \%\) O by mass.

(a) Combustion analysis of toluene, a common organic solvent, gives \(5.86 \mathrm{mg}\) of \(\mathrm{CO}_{2}\) and \(1.37 \mathrm{mg}\) of \(\mathrm{H}_{2} \mathrm{O}\). If the compound contains only carbon and hydrogen, what is its empirical formula? (b) Menthol, the substance we can smell in mentholated cough drops, is composed of \(\mathrm{C}, \mathrm{H},\) and \(\mathrm{O}\). A 0.1005 -g sample of menthol is combusted, producing \(0.2829 \mathrm{~g}\) of \(\mathrm{CO}_{2}\) and \(0.1159 \mathrm{~g}\) of \(\mathrm{H}_{2} \mathrm{O} .\) What is the empirical formula for menthol? If menthol has a molar mass of \(156 \mathrm{~g} / \mathrm{mol}\), what is its molecular formula?

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