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Chapter 5: Problem 103

Methanol (CH_l3 \(\mathrm{OH}\) ) can be produced by the following reaction: $$\mathrm{CO}(g)+2 \mathrm{H}_{2}(g) \longrightarrow \mathrm{CH}_{3} \mathrm{OH}(g)$$ Hydrogen at STP flows into a reactor at a rate of 16.0 L/min. Carbon monoxide at STP flows into the reactor at a rate of 25.0 L/min. If 5.30 g methanol is produced per minute, what is the percent yield of the reaction?

Short Answer

Expert verified
The percent yield of methanol in this reaction is 46.3%.

Step by step solution

01

Convert L/min flow rates of reactants into moles/min

At STP (standard temperature and pressure), 1 mole of an ideal gas occupies 22.4 L. We can use this information to convert the flow rates of hydrogen and carbon monoxide into moles/min: For hydrogen: \(\frac{16.0\,\text{L}}{\text{min}}\cdot\frac{1\,\text{mol}}{22.4\,\text{L}}=0.714\,\text{mol}\,\text{H}_{2}\,\text{min}^{-1}\) For carbon monoxide: \(\frac{25.0\,\text{L}}{\text{min}}\cdot\frac{1\,\text{mol}}{22.4\,\text{L}}=1.12\,\text{mol}\,\text{CO}\,\text{min}^{-1}\)
02

Determine the limiting reactant

To determine the limiting reactant, first find the ratio between the actual flow rates in mol/min and the stoichiometric coefficients from the balanced equation: For hydrogen: \(\frac{0.714\,\text{mol}\,\text{H}_{2}\,\text{min}^{-1}}{2}=0.357\) For carbon monoxide: \(\frac{1.12\,\text{mol}\,\text{CO}\,\text{min}^{-1}}{1}=1.12\) The lower ratio belongs to hydrogen, so it is the limiting reactant.
03

Calculate the theoretical yield of methanol

Now, we can use stoichiometry to determine the theoretical yield of methanol (CH₃OH). According to the balanced equation, 1 mol CO reacts with 2 mol H₂ to produce 1 mol CH₃OH. As hydrogen is the limiting reactant, we have: Theoretical yield of methanol = \(0.714\,\text{mol}\,\text{H}_{2}\,\text{min}^{-1}\cdot\frac{1\,\text{mol}\,\text{CH}_{3}\text{OH}}{2\,\text{mol}\,\text{H}_{2}}=0.357\,\text{mol}\,\text{CH}_{3}\text{OH}\,\text{min}^{-1}\) We need to convert this amount into grams: Theoretical yield of methanol = \(0.357\,\text{mol}\,\text{CH}_{3}\text{OH}\,\text{min}^{-1}\cdot\frac{32.04\,\text{g}}{1\,\text{mol}}=11.43\,\text{g}\,\text{CH}_{3}\text{OH}\,\text{min}^{-1}\)
04

Calculate the percent yield

Now, we can use the actual yield (5.30 g/min) and the theoretical yield (11.43 g/min) to calculate the percent yield: Percent yield = \(\frac{5.30\,\text{g}\,\text{CH}_{3}\text{OH}\,\text{min}^{-1}}{11.43\,\text{g}\,\text{CH}_{3}\text{OH}\,\text{min}^{-1}}\times100\%=46.3\%\) So the percent yield of methanol in this reaction is 46.3%.

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Most popular questions from this chapter

As \(\mathrm{NH}_{3}(g)\) is decomposed into nitrogen gas and hydrogen gas at constant pressure and temperature, the volume of the product gases collected is twice the volume of \(\mathrm{NH}_{3}\) reacted. Explain. As \(\mathrm{NH}_{3}(g)\) is decomposed into nitrogen gas and hydrogen gas at constant volume and temperature, the total pressure increases by some factor. Why the increase in pressure and by what factor does the total pressure increase when reactants are completely converted into products? How do the partial pressures of the product gases compare to each other and to the initial pressure of \(\mathrm{NH}_{3}?\)

Concentrated hydrogen peroxide solutions are explosively decomposed by traces of transition metal ions (such as \(\mathrm{Mn}\) or \(\mathbf{F} \mathrm{e}\)): $$2 \mathrm{H}_{2} \mathrm{O}_{2}(a q) \longrightarrow 2 \mathrm{H}_{2} \mathrm{O}(l)+\mathrm{O}_{2}(g)$$ What volume of pure \(\mathrm{O}_{2}(g),\) collected at \(27^{\circ} \mathrm{C}\) and 746 torr, would be generated by decomposition of 125 \(\mathrm{g}\) of a 50.0\(\% \mathrm{by}\) mass hydrogen peroxide solution? Ignore any water vapor that may be present.

In Example 5.11 of the text, the molar volume of \(\mathrm{N}_{2}(g)\) at \(\mathrm{STP}\) is given as 22.42 \(\mathrm{Lmol} \mathrm{N}_{2} .\) How is this number calculated? How does the molar volume of He(g) at \(\mathrm{STP}\) compare to the molar volume of \(\mathrm{N}_{2}(g)\) at \(\mathrm{STP}\) (assuming ideal gas behavior)? Is the molar volume of \(\mathrm{N}_{2}(g)\) at 1.000 atm and \(25.0^{\circ} \mathrm{C}\) equal to, less than, or greater than 22.42 \(\mathrm{L} / \mathrm{mol} ?\) Explain. Is the molar volume of \(\mathrm{N}_{2}(g)\) collected over water at a total pressure of 1.000 \(\mathrm{atm}\) and \(0.0^{\circ} \mathrm{C}\) equal to, less than, or greater than 22.42 \(\mathrm{L} / \mathrm{mol}\) ? Explain.

A bicycle tire is filled with air to a pressure of 75 psi at a temperature of \(19^{\circ} \mathrm{C}\) . Riding the bike on asphalt on a hot day increases the temperature of the tire to \(58^{\circ} \mathrm{C}\) . The volume of the tire increases by 4.0\(\%\) . What is the new pressure in the bicycle tire?

Consider three identical flasks filled with different gases. Flask \(\mathrm{A} : \mathrm{CO}\) at 760 torr and \(0^{\circ} \mathrm{C}\) Flask \(\mathrm{B} : \mathrm{N}_{2}\) at 250 torr and \(0^{\circ} \mathrm{C}\) Flask \(\mathrm{C} : \mathrm{H}_{2}\) at 100 torr and \(0^{\circ} \mathrm{C}\) a. In which flask will the molecules have the greatest average kinetic energy? b. In which flask will the molecules have the greatest average velocity?
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