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Chapter 3: Problem 103

The Hargreaves process is an industrial method for making sodium sulfate for use in papermaking. $$4 \mathrm{NaCl}+2 \mathrm{SO}_{2}+2 \mathrm{H}_{2} \mathrm{O}+\mathrm{O}_{2} \longrightarrow 2 \mathrm{Na}_{2} \mathrm{SO}_{4}+4 \mathrm{HCl}$$ (a) If you start with \(10 .\) mol of each reactant, determine which one limits the amount of \(\mathrm{Na}_{2} \mathrm{SO}_{4}\) produced. (b) Determine the amount of \(\mathrm{Na}_{2} \mathrm{SO}_{4}\) produced if you start with \(100 . \mathrm{g}\) of each reactant.

Short Answer

Expert verified
(a) NaCl limits Na鈧係O鈧� production, producing 5 mol. (b) 121.45 g of Na鈧係O鈧� is produced.

Step by step solution

01

Identify the Stoichiometric Coefficients

The balanced chemical equation is given as: \(4 \mathrm{NaCl}+2 \mathrm{SO}_{2}+2 \mathrm{H}_{2} \mathrm{O}+\mathrm{O}_{2} \rightarrow 2 \mathrm{Na}_{2} \mathrm{SO}_{4}+4 \mathrm{HCl}\). The stoichiometric coefficients for NaCl, SO鈧�, H鈧侽, O鈧�, and Na鈧係O鈧� are 4, 2, 2, 1, and 2, respectively.
02

Determine the Limiting Reactant (Moles)

Given 10 mol of each reactant: - For NaCl: \(\frac{10}{4} = 2.5\) equivalent reactions.- For SO鈧�: \(\frac{10}{2} = 5\) equivalent reactions.- For H鈧侽: \(\frac{10}{2} = 5\) equivalent reactions.- For O鈧�: \(\frac{10}{1} = 10\) equivalent reactions.Thus, NaCl is the limiting reactant because it allows the fewest complete reactions (2.5 reactions).
03

Calculate Na2SO4 Production from Limiting Reactant (Moles)

From the balanced equation, 4 mol of NaCl produces 2 mol of Na鈧係O鈧�. Therefore, using \(2.5\) complete reactions, the amount of Na鈧係O鈧� produced is: \(2.5\ times\ 2 = 5\) mol of Na鈧係O鈧�.
04

Convert Mass to Moles

Calculate the molar masses:- NaCl = 58.44 g/mol- SO鈧� = 64.07 g/mol- H鈧侽 = 18.02 g/mol- O鈧� = 32.00 g/molConvert each 100 g reactant to moles:- NaCl: \(100 \div 58.44 = 1.71\) mol- SO鈧�: \(100 \div 64.07 = 1.56\) mol- H鈧侽: \(100 \div 18.02 = 5.55\) mol- O鈧�: \(100 \div 32.00 = 3.13\) mol
05

Identify Limiting Reactant (Mass)

Using the converted moles and stoichiometric ratios:- For NaCl: \(\frac{1.71}{4} = 0.4275\) equivalent reactions- For SO鈧�: \(\frac{1.56}{2} = 0.78\) equivalent reactions- For H鈧侽: \(\frac{5.55}{2} = 2.775\) equivalent reactions- For O鈧�: \(\frac{3.13}{1} = 3.13\) equivalent reactionsNaCl is the limiting reactant due to the lowest value of 0.4275 reactions.
06

Calculate Na2SO4 Production from Limiting Reactant (Mass)

Using 0.4275 reactions and finding Na鈧係O鈧� produced:Given that 4 mol NaCl produces 2 mol Na鈧係O鈧�:\(0.4275 \times 2 = 0.855\) mol Na鈧係O鈧� produced.Molar mass of Na鈧係O鈧� is 142.04 g/mol, thus:\[0.855 \times 142.04 = 121.45\] grams of Na鈧係O鈧�.

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Key Concepts

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

Stoichiometry
Stoichiometry is a key concept in chemistry that involves the calculation of reactants and products in chemical reactions. It uses the balanced chemical equation to determine the proportions needed for the reaction to occur.
To begin with, a balanced equation reflects the conservation of mass, where the number of atoms of each element is the same on both sides of the equation. This includes recognizing stoichiometric coefficients, which are the numbers before each compound.
  • For the equation: \(4 \mathrm{NaCl} + 2 \mathrm{SO}_{2} + 2 \mathrm{H}_{2} \mathrm{O} + \mathrm{O}_{2} \rightarrow 2 \mathrm{Na}_{2} \mathrm{SO}_{4} + 4 \mathrm{HCl}\), these coefficients tell us that 4 moles of \(\mathrm{NaCl}\) will produce 2 moles of \(\mathrm{Na}_2\mathrm{SO}_4\).
  • When solving problems involving stoichiometry, we assess the given quantities to establish which reactant limits the reaction. In this exercise, sodium chloride (\(\mathrm{NaCl}\)) is identified as the limiting reactant when starting with 10 moles of each reactant.
By using stoichiometry, you can calculate the theoretical maximum amount of products that can be formed from specified amounts of reactants.
Sodium Sulfate Production
Sodium sulfate (\(\mathrm{Na}_2\mathrm{SO}_4\)) is an important compound with various industrial uses, such as in the Hargreaves process for papermaking.
In this process, sodium chloride, sulfur dioxide, water, and oxygen react to form sodium sulfate and hydrochloric acid.
Understanding how much sodium sulfate can be made requires determining the limiting reactant since it's the component that will run out first, stopping the production.
  • For instance, in part (a) of the exercise where each reactant starts at 10 moles, sodium chloride limits the reaction to produce \(5\) moles of \(\mathrm{Na}_2\mathrm{SO}_4\).
  • Moreover, when starting with \(100\, \mathrm{g}\) of each reactant, the calculation shows \(121.45\, \mathrm{g}\) of \(\mathrm{Na}_2\mathrm{SO}_4\) is produced as per part (b), due to sodium chloride again acting as the limiting reactant.
This equation demonstrates how knowing the chemical process in detail allows industries to optimize the amounts of product obtained from given reactants.
Molar Mass Calculation
Calculating molar masses is crucial when converting between moles and grams for substances in chemical reactions.
The molar mass is essentially the mass of one mole of a given substance, typically reported in grams per mole (g/mol).
  • In the context of the exercise, some of the molar masses include: the molar mass of \(\mathrm{NaCl}\) is \(58.44\, \mathrm{g/mol}\), \(\mathrm{SO}_2\) is \(64.07\, \mathrm{g/mol}\), \(\mathrm{H}_2\mathrm{O}\) is \(18.02\, \mathrm{g/mol}\), and \(\mathrm{O}_2\) is \(32.00\, \mathrm{g/mol}\).
  • These values are used to convert between grams and moles, which is necessary to find out the number of moles of reactants.
    For example, in step (4), \(100\, \mathrm{g}\) of \(\mathrm{NaCl}\) is equivalent to \(1.71\) moles because \(100\, \mathrm{g} \div 58.44\, \mathrm{g/mol} \approx 1.71\).
Once the moles are determined for each reactant, it becomes straightforward to compute how much product can be formed by comparing these to the stoichiometric coefficients from the balanced equation. This is a fundamental technique in chemistry to translate laboratory results to scalable industrial processes.

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Most popular questions from this chapter

You mix \(25.0 \mathrm{~mL}\) of \(0.234-\mathrm{M} \mathrm{FeCl}_{3}\) solution with \(42.5 \mathrm{~mL}\) of \(0.453-\mathrm{M} \mathrm{NaOH}\) (a) Calculate the maximum mass, in grams, of \(\mathrm{Fe}(\mathrm{OH})_{3}\) that will precipitate. (b) Determine which reactant is in excess. (c) Calculate the concentration of the excess reactant remaining in solution after the maximum mass of \(\mathrm{Fe}(\mathrm{OH})_{3}\) has precipitated.

Chlorine has several oxidation states. (a) Determine the oxidation state of chlorine in \(\mathrm{Cl}_{2}, \mathrm{HOCl}, \mathrm{ClO}_{2}, \mathrm{NH}_{4} \mathrm{Cl},\) and \(\mathrm{NaCl}\). (b) Chloride ion is the most stable form. Predict which of the other substances given in part (a) is an oxidizing or a reducing agent and rank them in decreasing order of their oxidizing strengths.

Iron oxide can be reduced to the metal as follows: $$\mathrm{Fe}_{2} \mathrm{O}_{3}(\mathrm{~s})+3 \mathrm{CO}(\mathrm{g}) \longrightarrow 2 \mathrm{Fe}(\mathrm{s})+3 \mathrm{CO}_{2}(\mathrm{~g})$$ Calculate the mass of iron that can be obtained from \(1.00 \mathrm{~kg}\) of the iron oxide. If \(654 \mathrm{~g}\) Fe was obtained from the reaction, calculate the percent yield.

Sulfur can exist in many oxidation states. Determine the oxidation state of \(\mathrm{S}\) in each species. (a) \(\mathrm{H}_{2} \mathrm{~S}\) (b) \(\mathrm{S}_{8}\) (c) \(\mathrm{SCl}_{2}\) (d) \(\mathrm{SO}_{3}^{2-}\) (e) \(\mathrm{K}_{2} \mathrm{SO}\)

Classify each of these as an acid or a base. Which are strong and which are weak? What ions are produced when each is dissolved in water? (a) \(\mathrm{HNO}_{3}\) (b) \(\mathrm{Ca}(\mathrm{OH})_{2}\) (c) \(\mathrm{NH}_{3}\) (d) \(\mathrm{H}_{3} \mathrm{PO}_{4}\) (e) \(\mathrm{KOH}\) (f) \(\mathrm{CH}_{3} \mathrm{COOH}\)
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