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A chemical plant uses electrical energy to decompose aqueous solutions of 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 \(1.5 \times 10^{6} \mathrm{kg}\left(1500\) metric tons) of \(\mathrm{Cl}_{2}\right.\) daily, estimate the quantities of \(\mathrm{H}_{2}\) and \(\mathrm{NaOH}\) produced.

Short Answer

Expert verified
The chemical plant produces approximately \(4.28 \times 10^{4} \thinspace \mathrm{kg}\) of hydrogen gas (H₂) and \(1.69 \times 10^{6} \thinspace \mathrm{kg}\) of sodium hydroxide (NaOH) daily, along with the given \(1.5 \times 10^6 \thinspace \mathrm{kg}\) of chlorine gas (Cl₂).

Step by step solution

01

Write down the balanced chemical equation.

The balanced chemical equation for the decomposition of aqueous solutions of NaCl is given as: $$ 2 \mathrm{NaCl}(a q)+2 \mathrm{H}_{2} \mathrm{O}(l) \longrightarrow 2 \mathrm{NaOH}(a q)+\mathrm{H}_{2}(g)+\mathrm{Cl}_{2}(g) $$
02

Calculate moles of Cl₂ produced daily.

We are given that the plant produces \(1.5 \times 10^{6} \mathrm{kg}\) of Cl₂ daily. To find the moles of Cl₂, we can use the following formula: $$ \text{moles of Cl₂} = \frac{\text{mass of Cl₂}}{\text{molar mass of Cl₂}} $$ The molar mass of Cl₂ is \(2 \times 35.45 \thinspace \mathrm{g/mol}\) (since there are two chlorine atoms, each with a molar mass of \(35.45 \thinspace \mathrm{g/mol}\)). First, convert the mass of Cl₂ from kg to g: $$ 1.5 \times 10^{6} \thinspace \mathrm{kg} \times \frac{1000 \thinspace \mathrm{g}}{1 \thinspace \mathrm{kg}} = 1.5 \times 10^{9} \thinspace \mathrm{g} $$ Next, calculate the moles of Cl₂ produced daily: $$ \text{moles of Cl₂} = \frac{1.5 \times 10^{9} \thinspace \mathrm{g}}{2 \times 35.45 \thinspace \mathrm{g/mol}} = \frac{1.5 \times 10^{9} \thinspace \mathrm{g}}{70.9 \thinspace \mathrm{g/mol}} = 2.12 \times 10^{7} \thinspace \mathrm{mol} $$
03

Calculate moles of H₂ and NaOH produced daily.

From the balanced equation, we can see that for each mole of Cl₂ produced, one mole of H₂ is produced, and two moles of NaOH are produced. Therefore: Moles of H₂ produced daily = Moles of Cl₂ produced daily = \(2.12 \times 10^{7} \thinspace \mathrm{mol}\) Moles of NaOH produced daily = \(2 \times\) Moles of Cl₂ produced daily = \(2 \times 2.12 \times 10^{7} \thinspace \mathrm{mol} = 4.24 \times 10^{7} \thinspace \mathrm{mol}\)
04

Calculate mass of H₂ and NaOH produced daily.

Now we can calculate the mass of H₂ and NaOH produced daily using their respective molar masses: Molar mass of H₂ = \(2 \times 1.01 \thinspace \mathrm{g/mol}\) (since there are two hydrogen atoms, each with a molar mass of \(1.01 \thinspace \mathrm{g/mol}\)) Molar mass of NaOH = \(22.99 \thinspace \mathrm{g/mol} + 15.999 \thinspace \mathrm{g/mol} + 1.01 \thinspace \mathrm{g/mol}\) (Na, O, and H) Mass of H₂ produced daily: $$ (2.12 \times 10^{7} \thinspace \mathrm{mol}) \times (2.02 \thinspace \mathrm{g/mol}) = 4.28 \times 10^{7} \thinspace \mathrm{g} = 4.28 \times 10^{4} \thinspace \mathrm{kg} $$ Mass of NaOH produced daily: $$ (4.24 \times 10^{7} \thinspace \mathrm{mol}) \times (39.998\thinspace \mathrm{g/mol}) = 1.69 \times 10^{9} \thinspace \mathrm{g} = 1.69 \times 10^{6} \thinspace \mathrm{kg} $$
05

Conclusion

The chemical plant produces approximately \(4.28 \times 10^{4} \thinspace \mathrm{kg}\) of hydrogen gas (H₂) and \(1.69 \times 10^{6} \thinspace \mathrm{kg}\) of sodium hydroxide (NaOH) daily, along with the given \(1.5 \times 10^6 \thinspace \mathrm{kg}\) of chlorine gas (Cl₂).

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Key Concepts

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

Chemical Reactions
In a chemical reaction, substances called reactants are transformed into different substances known as products. This transformation involves breaking and forming of chemical bonds. For our exercise, the chemical plant uses electrical energy to decompose aqueous NaCl, a process where you start with NaCl and water (H₂O) as the reactants, and produce NaOH, H₂, and Cl₂ as the products. This type of reaction is known as electrolysis whereby electricity prompts the decomposition of chemicals.
Molar Mass
The concept of molar mass is essential to convert between the mass of a substance and the number of moles. It represents the mass of one mole of a given substance, expressed in grams per mole (g/mol). For Cl₂, each chlorine atom has a molar mass of 35.45 g/mol. Since Cl₂ consists of two chlorine atoms, its molar mass will be \(2 \times 35.45\ g/mol = 70.9\ g/mol\).
Understanding molar mass enables us to calculate how many moles are present in a certain mass of a substance, which is a crucial step in stoichiometric calculations of chemical reactions.
Balanced Chemical Equations
A balanced chemical equation accurately represents the conservation of mass in a chemical reaction, ensuring that the same number of each type of atom appears on both sides of the equation. In the given reaction:\[2 \mathrm{NaCl}(aq) + 2 \mathrm{H}_2\mathrm{O}(l) \rightarrow 2 \mathrm{NaOH}(aq) + \mathrm{H}_2(g) + \mathrm{Cl}_2(g)\]we see that the equation is balanced. Each element has the same number of atoms on both sides, for instance, two chlorine, two sodium, and four hydrogen atoms in both reactants and products. Balancing the equation is fundamental to understanding the stoichiometric relationships between reactants and products, enabling precise calculations in reactions.
Mass Calculations
Mass calculations involve determining the amount of product formed from a given amount of reactants, or vice versa. This concept is crucial in chemical reactions to understand the full scope of the transformation process. To find the mass of H₂ and NaOH produced daily, we use their respective molar masses:
  • For H₂: \(2.02\ g/mol\) (due to two hydrogen atoms each at \(1.01\ g/mol\)).
  • For NaOH: \(39.998\ g/mol\), calculated from the individual molar masses of Na, O, and H.
Calculating mass involves multiplying the number of moles by the molar mass to convert between moles and grams. For example, the plant produces \(4.28 \times 10^4\ kg\) of hydrogen and \(1.69 \times 10^6\ kg\) of NaOH daily, providing a clear understanding of the output from the chemical reaction process.

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Most popular questions from this chapter

When benzene \(\left(\mathrm{C}_{6} \mathrm{H}_{6}\right)\) reacts with bromine \(\left(\mathrm{Br}_{2}\right),\) bromobenzene \(\left(\mathrm{C}_{6} \mathrm{H}_{5} \mathrm{Br}\right)\) is obtained: $$ \mathrm{C}_{6} \mathrm{H}_{6}+\mathrm{Br}_{2} \longrightarrow \mathrm{C}_{6} \mathrm{H}_{5} \mathrm{Br}+\mathrm{HBr} $$ (a) When 30.0 g of benzene reacts with 65.0 g of bromine, what is the theoretical yield of bromobenzene? (b) If the actual yield of bromobenzene is \(42.3 \mathrm{g},\) what is the percentage yield?

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When a mixture of 10.0 g of acetylene \(\left(\mathrm{C}_{2} \mathrm{H}_{2}\right)\) and 10.0 \(\mathrm{g}\) of oxygen \(\left(\mathrm{O}_{2}\right)\) is ignited, the resulting combustion reaction produces \(\mathrm{CO}_{2}\) and \(\mathrm{H}_{2} \mathrm{O}\) . (a) Write the balanced chemical equation for this reaction. (b) Which is the limiting reactant? (c) How many grams of \(\mathrm{C}_{2} \mathrm{H}_{2}, \mathrm{O}_{2}, \mathrm{CO}_{2},\) and \(\mathrm{H}_{2} \mathrm{O}\) are present after the reaction is complete?

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: $$ \begin{array}{l}{2 \mathrm{KClO}_{3}(s) \longrightarrow 2 \mathrm{KCl}(s)+3 \mathrm{O}_{2}(g)} \\ {2 \mathrm{KHCO}_{3}(s) \longrightarrow \mathrm{K}_{2} \mathrm{O}(s)+\mathrm{H}_{2} \mathrm{O}(g)+2 \mathrm{CO}_{2}(g)} \\\ {\mathrm{K}_{2} \mathrm{CO}_{3}(s) \longrightarrow \mathrm{K}_{2} \mathrm{O}(s)+\mathrm{CO}_{2}(g)}\end{array} $$ The KCl does not react under the conditions of the reaction. If 100.0 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.)

Solutions of sodium carbonate and silver nitrate react to form solid silver carbonate and a solution of sodium nitrate. A solution containing 3.50 \(\mathrm{g}\) of sodium carbonate is mixed with one containing 5.00 \(\mathrm{g}\) of silver nitrate. How many grams of sodium carbonate, silver nitrate, silver carbonate, and sodium nitrate are present after the reaction is complete?

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