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Chapter 8: Problem 95

Methanol \(\left(\mathrm{CH}_{3} \mathrm{OH}\right)\) 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 \(16.0 \mathrm{L} / \mathrm{min.}\) Carbon monoxide at STP flows into the reactor at a rate of \(25.0 \mathrm{L} / \mathrm{min.}\). If \(5.30\) \(\mathrm{g}\) methanol is produced per minute, what is the percent yield of the reaction?

Short Answer

Expert verified
The percent yield for the given reaction can be calculated using the following steps: 1. Calculate moles of CO and H₂ using the given volumes and STP conditions. 2. Determine the limiting reactant by comparing mole ratios. 3. Calculate the theoretical mass of Methanol based on the limiting reactant. 4. Calculate the percent yield using the formula: \(\text{Percent Yield} = \frac{\text{Actual Mass of CH}_{3}\text{OH}}{\text{Theoretical Mass of CH}_{3}\text{OH}} * 100\) Plug in the values and calculate the percent yield.

Step by step solution

01

Calculate moles of the given reactants

To calculate the theoretical Methanol production, we first need to convert the given volumes of CO and H₂ at STP into moles. We can use the equation: n = PV/RT At STP (Standard Temperature and Pressure), R = 0.0821 L atm K⁻¹ mol⁻¹, T = 273.15 K, and P = 1 atm. Thus, equation becomes: n = V/R T For the given volumes of CO and H₂, plug into the equation and calculate their moles: Moles of CO = \(25.0 / (0.0821 * 273.15)\) Moles of H₂ = \(16.0 / (0.0821 * 273.15)\)
02

Determine the limiting reactant

Compare the mole ratio of CO and H₂ according to the balanced equation (ratio is 1:2). Then, calculate the limiting reactant and how many moles of Methanol can be produced as per the stoichiometric ratio. For the given moles of CO and H₂, calculate the ratios: Ratio_CO = Moles of CO Ratio_H₂ = Moles of H₂ / 2 Compare these ratios and determine the limiting reactant and calculate the number of moles of Methanol that can be theoretically produced: If Ratio_CO < Ratio_H₂, then CO is the limiting reactant and moles of CH₃OH = moles of CO If Ratio_H₂ < Ratio_CO, then H₂ is the limiting reactant and moles of CH₃OH = moles of H₂ / 2
03

Calculate the theoretical mass of Methanol

Now, convert the moles of Methanol we determined based on limiting reactant to mass using molecular weight (32 g/mol). Mass of CH₃OH (theoretical) = Moles of CH₃OH * 32 g/mol
04

Calculate the percent yield

In the end, calculate the percent yield using the formula: Percent Yield = \(\frac{Actual \ Mass \ of \ CH_{3}OH}{Theoretical \ Mass \ of \ CH_{3}OH} * 100\) Percentage yield = \(\frac{5.30}{Mass \ of \ CH_{3}OH \ (theoretical)} * 100\) This will give us the percent yield for the given reaction.

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Most popular questions from this chapter

Helium is collected over water at \(25^{\circ} \mathrm{C}\) and 1.00 atm total pressure. What total volume of gas must be collected to obtain \(0.586 \mathrm{g}\) helium? (At \(25^{\circ} \mathrm{C}\) the vapor pressure of water is 23.8 torr.)

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