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Chapter 3: Problem 98

A chemical plant uses electrical energy to decompose aqueous solutions of \(\mathrm{NaCl}\) to give \(\mathrm{Cl}_{2}, \mathrm{H}_{2}\), and \(\mathrm{NaOH}\) : $$ \begin{aligned} \mathrm{NaCl}(a q)+2 \mathrm{H}_{2} \mathrm{O}(l) \longrightarrow & \\ & 2 \mathrm{NaOH}(a q)+\mathrm{H}_{2}(g)+\mathrm{Cl}_{2}(g) \end{aligned} $$ 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 estimated quantities of H鈧� and NaOH produced when 1.5 x 10鈦� kg of Cl鈧� is produced are approximately 42,690 kg of H鈧� and 1,694,000 kg of NaOH.

Step by step solution

01

Balance the chemical equation

In order to solve this problem, we need to have a balanced chemical equation. The given equation is already balanced, as shown below: \(2 NaCl(aq) + 2 H_2O(l) \longrightarrow 2 NaOH(aq) + H_2(g) + Cl_2(g)\)
02

Convert the mass of Cl鈧� into moles

We are given the mass of Cl鈧� produced (1.5 x 10鈦� kg). To make use of stoichiometry, we need to convert the mass into moles. We can do that using the molar mass of Cl鈧�, which is 35.45 x 2 = 70.9 g/mol: moles of Cl鈧� = \(\frac{mass}{molar~mass}\) = \(\frac{1.5\times10^{6}~kg}{70.9~g/mol}\) As we're dealing with units involving kg and g: moles of Cl鈧� = \(\frac{1.5\times10^{6}~kg \times 10^{3}~g/kg}{70.9~g/mol}\) moles of Cl鈧� = \(2.118 \times 10^{7}~mol\)
03

Calculate the moles of H鈧� produced

Using stoichiometry and the balanced chemical equation, we can determine the moles of H鈧� produced by comparing the stoichiometric coefficients: \(2 NaCl(aq) + 2 H_2O(l) \longrightarrow 2 NaOH(aq) + H_2(g) + Cl_2(g)\) From the balanced equation, we can observe that 1 mole of H鈧� is produced for every 1 mole of Cl鈧�. Therefore, the moles of H鈧� produced is equal to the moles of Cl鈧�: moles of H鈧� = moles of Cl鈧� = \(2.118 \times 10^{7}~mol\)
04

Calculate the moles of NaOH produced

Similarly, we can determine the moles of NaOH produced using stoichiometry. From the balanced equation, we can observe that 2 moles of NaOH are produced for every 1 mole of Cl鈧�. Therefore, we can calculate the moles of NaOH produced: moles of NaOH = 2 x moles of Cl鈧� = 2 x \(2.118 \times 10^{7}~mol\) moles of NaOH = \(4.236 \times 10^{7}~mol\)
05

Convert the moles of H鈧� and NaOH back to mass

Finally, to obtain the quantities of H鈧� and NaOH produced, we need to convert the moles back into mass using their respective molar masses. The molar mass of H鈧� is 2.016 g/mol, and that of NaOH is 22.99 + 15.999 + 1.008 = 39.998 g/mol. So, mass of H鈧� produced = moles of H鈧� x molar mass of H鈧� mass of H鈧� produced = \(2.118 \times 10^{7}~mol \times 2.016~g/mol\) mass of H鈧� produced = \(4.269 \times 10^{7}~g\) or \(4.269 \times 10^{4}~kg\) mass of NaOH produced = moles of NaOH x molar mass of NaOH mass of NaOH produced = \(4.236 \times 10^{7}~mol \times 39.998~g/mol\) mass of NaOH produced = \(1.694 \times 10^{9}~g\) or \(1.694 \times 10^{6}~kg\) So, the estimated quantities of H鈧� and NaOH produced are approximately 42,690 kg of H鈧� and 1,694,000 kg of NaOH.

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Key Concepts

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

Stoichiometry
Stoichiometry is a fundamental concept in chemistry that deals with the quantitative relationships between reactants and products in a chemical reaction. Understanding stoichiometry is essential when you're trying to predict how much product will be formed from given quantities of reactants. This is crucial in industrial applications, like the chemical plant in the problem, where precise amounts of substances are required for production processes.

In this exercise, stoichiometry helps us determine the amount of hydrogen and sodium hydroxide produced from the decomposition of sodium chloride. By looking at the balanced chemical equation, you can see that the coefficients give us the mole ratios of the reactants to products. For example:

  • 1 mole of \( ext{Cl}_2\) is produced alongside 1 mole of \( ext{H}_2\)
  • 2 moles of \( ext{NaOH}\) are produced for each mole of \( ext{Cl}_2\)
These ratios allow for the calculation of product masses from given reactant amounts, ensuring that the chemical process is efficient and predictable.
Chemical Reactions
Understanding chemical reactions is crucial as they describe how different substances interact to form new compounds. In the context of electrolysis, a chemical reaction, electrical energy is used to drive a non-spontaneous reaction.

In our particular exercise, the chemical reaction is set up to decompose sodium chloride \( ext{NaCl}\) in solution. This decomposition yields \( ext{NaOH}\), \( ext{H}_2\), and \( ext{Cl}_2\). This reaction not only involves changes in the physical states of matter (aqueous to gaseous) but also electron transfer, as indicated in redox processes in electrolysis.

By balancing the chemical equation, we ensure that the law of conservation of mass is satisfied. This law states that matter cannot be created or destroyed in a chemical reaction, ensuring we have the same number and kind of atoms on both sides of the equation.
Molar Mass Calculation
Molar mass is a key concept in chemistry, defined as the mass of one mole of a substance. It serves as a bridge between the micro (molecular) world and the macro (numerical) world, allowing for conversion between mass and amount in moles.

In this problem, we utilized the molar masses of chlorine gas, hydrogen gas, and sodium hydroxide to carry out necessary calculations. For example, the molar mass of \( ext{Cl}_2\) is determined by doubling the atomic mass of chlorine (35.45 g/mol), resulting in 70.9 g/mol. Similarly, the mass of \( ext{NaOH}\) is derived from adding the atomic masses of sodium (22.99 g/mol), oxygen (15.999 g/mol), and hydrogen (1.008 g/mol), totaling to approximately 39.998 g/mol.

This knowledge is applied when converting between mass and moles, vital in calculating the quantities of substances produced in the chemical process. Understanding how to accurately calculate and use molar mass is fundamental for predicting yields and ensuring the precision of chemical manufacturing.

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Most popular questions from this chapter

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A compound whose empirical formula is \(\mathrm{XF}_{3}\) consists of \(65 \%\) F by mass. What is the atomic mass of X?

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(a) Determine the chemical formula of the product formed when the metallic element aluminum combines with the nonmetallic element bromine, \(\mathrm{Br}_{2}\). Write the balanced chemical equation for the reaction. (b) What products form when a compound containing \(\mathrm{C}, \mathrm{H},\) and \(\mathrm{O}\) is completely combusted in air? Write a balanced chemical equation for the combustion of acetone, \(\mathrm{C}_{3} \mathrm{H}_{6} \mathrm{O}(l),\) in air.

Without doing any detailed calculations (but using a periodic table to give atomic weights), rank the following samples in order of increasing number of atoms: \(9.0 \times 10^{23}\) molecules of \(\mathrm{H}_{2} \mathrm{O}_{2}, 2.0 \mathrm{~mol} \mathrm{CH}_{4}, 16 \mathrm{~g} \mathrm{O}_{2}\)
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