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Chapter 4: Problem 67

A side reaction in the manufacture of rayon from wood pulp is \(3 \mathrm{CS}_{2}+6 \mathrm{NaOH} \longrightarrow 2 \mathrm{Na}_{2} \mathrm{CS}_{3}+\mathrm{Na}_{2} \mathrm{CO}_{3}+3 \mathrm{H}_{2} \mathrm{O}\) How many grams of \(\mathrm{Na}_{2} \mathrm{CS}_{3}\) are produced in the reaction of \(92.5 \mathrm{mL}\) of liquid \(\mathrm{CS}_{2}(d=1.26 \mathrm{g} / \mathrm{mL})\) and 2.78 mol NaOH?

Short Answer

Expert verified
The mass of \(\mathrm{Na}_{2}\mathrm{CS}_{3}\) produced in the reaction is approximately 146.34 g.

Step by step solution

01

Convert volume to mass of CS2

First, convert the volume of \(\mathrm{CS}_{2}\) to mass by using the given density (d = 1.26 g/mL). The formula to calculate mass from volume and density is: \[ \text{Mass} = \text{Density} * \text{Volume} \]. Convert 92.5 mL of \(\mathrm{CS}_{2}\) to grams which is: \( 1.26 \, \mathrm{g/mL} * 92.5 \, \mathrm{mL} = 116.55 \, \mathrm{g} \) of \(\mathrm{CS}_{2}\).
02

Convert mass to moles of CS2

To solve this problem, we must deal with moles. The formula to find moles from mass and molar mass is: \[ \text{Moles} = \frac{\text{Mass}}{\text{Molar mass}} \]. The molar mass of \(\mathrm{CS}_{2}\) is 76.141 g/mol. Substituting the given values, we derive that there are \( \frac{116.55 \, \mathrm{g}}{76.141 \, \mathrm{g/mol}} = 1.530 \, \mathrm{mol} \) of \(\mathrm{CS}_{2}\).
03

Use stoichiometric ratios

According to the balanced chemical reaction, 3 moles of \(\mathrm{CS}_{2}\) produces 2 moles of \(\mathrm{Na}_{2}\mathrm{CS}_{3}\). Thus, we will calculate the moles of \(\mathrm{Na}_{2}\mathrm{CS}_{3}\) that could be produced given the moles of \(\mathrm{CS}_{2}\) and \(\mathrm{NaOH}\) according to these stoichiometric ratios. Based on \(\mathrm{CS}_{2}\), \(\frac{1.530 \, \mathrm{mol} \, \mathrm{CS}_{2}}{3} * 2 = 1.020 \, \mathrm{mol} \, \mathrm{Na}_{2}\mathrm{CS}_{3}\) can be theoretically produced. According to \(\mathrm{NaOH}\), \(\frac{2.78 \, \mathrm{mol} \, \mathrm{NaOH}}{6} * 2 = 0.926 \, \mathrm{mol} \, \mathrm{Na}_{2}\mathrm{CS}_{3}\) can be theoretically produced.
04

Identify the limiting reactant and calculate the mass of sodium thiocarbonate

Since fewer moles of \(\mathrm{Na}_{2}\mathrm{CS}_{3}\) can be formed from \(\mathrm{NaOH}\), NaOH is the limiting reactant. We calculate the mass of \(\mathrm{Na}_{2}\mathrm{CS}_{3}\) that can be formed from this amount of \(\mathrm{NaOH}\) using this formula: \[ \text{Mass} = \text{Moles} * \text{Molar mass} \]. The molar mass of \(\mathrm{Na}_{2}\mathrm{CS}_{3}\) is 158.11 g/mol. Substituting the given values, we derive that \( 0.926 \, \mathrm{mol} * 158.11 \, \mathrm{g/mol} = 146.34 \, \mathrm{g} \, \mathrm{Na}_{2}\mathrm{CS}_{3} \).
05

Perform check

To make sure everything has been calculated correctly, check the relation between the moles of \(\mathrm{NaOH}\) and the moles of \(\mathrm{CS}_{2}\). As seen from the chemical equation, the mole ratio \(\mathrm{NaOH} : \mathrm{CS}_{2}\) should be 6:3 or 2:1. The given moles of \(\mathrm{NaOH}\) is 2.78 mol, and the moles of \(\mathrm{CS}_{2}\) is 1.530 mol. These values do indeed respect the 2:1 ratio, so our answer is most likely correct.

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Most popular questions from this chapter

Solid calcium carbonate, \(\mathrm{CaCO}_{3}(\mathrm{s}),\) reacts with \(\mathrm{HCl}(\mathrm{aq})\) to form \(\mathrm{H}_{2} \mathrm{O}, \mathrm{CaCl}_{2}(\mathrm{aq}),\) and \(\mathrm{CO}_{2}(\mathrm{g}) .\) If a \(45.0 \mathrm{g}\) sample of \(\mathrm{CaCO}_{3}(\mathrm{s})\) is added to \(1.25 \mathrm{L}\) of \(\mathrm{HCl}(\mathrm{aq})\) that is \(25.7 \% \mathrm{HCl}\) by mass \((d=1.13 \mathrm{g} / \mathrm{mL})\) what will be the molarity of \(\mathrm{HCl}\) in the solution after the reaction is completed? Assume that the solution volume remains constant.

A 0.696 mol sample of \(\mathrm{Cu}\) is added to \(136 \mathrm{mL}\) of \(6.0 \mathrm{M}\) HNO \(_{3}\). Assuming the following reaction is the only one that occurs, will the Cu react completely? $$\begin{aligned} 3 \mathrm{Cu}(\mathrm{s})+8 \mathrm{HNO}_{3}(\mathrm{aq}) & \longrightarrow 3 \mathrm{Cu}\left(\mathrm{NO}_{3}\right)_{2}(\mathrm{aq}) +4 \mathrm{H}_{2} \mathrm{O}(\mathrm{l})+2 \mathrm{NO}(\mathrm{g}) \end{aligned}$$

Ammonia can be generated by heating together the solids \(\mathrm{NH}_{4} \mathrm{Cl}\) and \(\mathrm{Ca}(\mathrm{OH})_{2} . \mathrm{CaCl}_{2}\) and \(\mathrm{H}_{2} \mathrm{O}\) are also formed. (a) If a mixture containing \(33.0 \mathrm{g}\) each of \(\mathrm{NH}_{4} \mathrm{Cl}\) and \(\mathrm{Ca}(\mathrm{OH})_{2}\) is heated, how many grams of \(\mathrm{NH}_{3}\) will form? (b) Which reactant remains in excess, and in what mass?

The incomplete combustion of gasoline produces \(\mathrm{CO}(\mathrm{g})\) as well as \(\mathrm{CO}_{2}(\mathrm{g}) .\) Write an equation for \((\mathrm{a})\) the complete combustion of the gasoline component octane, \(\mathrm{C}_{8} \mathrm{H}_{18}(\mathrm{l}),\) and \((\mathrm{b})\) incomplete combustion of octane with \(25 \%\) of the carbon appearing as \(\mathrm{CO}(\mathrm{g})\)

Cryolite, \(\mathrm{Na}_{3} \mathrm{AlF}_{6^{\prime}}\) is an important industrial reagent. It is made by the reaction below. $$\begin{array}{r} \mathrm{Al}_{2} \mathrm{O}_{3}(\mathrm{s})+6 \mathrm{NaOH}(\mathrm{aq})+12 \mathrm{HF}(\mathrm{g}) \longrightarrow 2 \mathrm{Na}_{3} \mathrm{AlF}_{6}(\mathrm{s})+9 \mathrm{H}_{2} \mathrm{O}(\mathrm{l}) \end{array}$$ In an experiment, \(7.81 \mathrm{g} \mathrm{Al}_{2} \mathrm{O}_{3}\) and excess \(\mathrm{HF}(\mathrm{g})\) were dissolved in 3.50 L of 0.141 M NaOH. If 28.2 g \(\mathrm{Na}_{3} \mathrm{AlF}_{6}\) was obtained, then what is the percent yield for this experiment?
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