Exercise 5.80 Enhanced - with Feedback Part A Carbon monoxide gas reacts with hy

Question

Exercise 5.80 Enhanced - with Feedback Part A Carbon monoxide gas reacts with hydrogen gas to form methanol via the following reaction ldentify the limiting reactant and determine the theoretical yield of methanol in grams Express your answer with the appropriate units CO(g) + 2H2 (g) CH3OH(g) A 1.70 L reaction vessel, initially at 305 K, contains carbon monoxide gas at a partial pressure of 232 mmHg and hydrogen gas at a partial pressure of 362 mmHg Value Units You may want to reference (mpage:23-223) Section 5.7 while completing this problem. Submit My Answers Give Up Provide Feedback Continue

Explanation / Answer

Ans. Given, Volume of reaction vessel, V= 1.70 L

            Temperature, T = 305 K

            Partial pressure of CO = 232 mmHg= (232 / 760) atm = 0.305263 atm

            Partial pressure of H2 = 362 mmHg= (362 / 760) atm = 0.476316 atm

# Calculate number of moles of CO:

Using Ideal gas equation:    PV = nRT      - equation 1

            Where, P = pressure in atm

            V = volume in L                   

            n = number of moles

            R = universal gas constant= 0.0821 atm L mol-1K-1

            T = absolute temperature (in K) = (0C + 273.15) K

Putting the values in equation 1-

            0.305263 atm x 1.70 L = n x (0.0821 atm L mol-1K-1) x 305 K

            Or, n = 0.020724 mol

# Calculate number of moles of H2:

Putting the values in equation 1-

            0.476316 atm x 1.70 L = n x (0.0821 atm L mol-1K-1) x 305 K

            Or, n = 0.032337 mol

# Determine limiting reactant:

In the balanced, the theoretical molar ration of reactants =

                        CO : H2 = 1 mol : 2 mol = 1: 2

Experimental molar ratio of reactants = CO : H2 = 0.020724 mol / 0.032337 mol = 1 : 1.56

Comparing the theoretical and experimental molar ratios of recants, the experimental moles of H2 are less than theoretical value 2.0 mol while keeping the moles of CO constant at 1.0.

Therefore,

            H2 is the limiting reactant.

# Theoretical yield of methanol:

Formation of product follows the stoichiometry of limiting reactant.

In the balanced reaction, 2 mol H2 forms 1 mol methanol.

So,

            Theoretical moles of methanol formed = (1/2) x moles of H2 consumed

                                                            = (1/2) x 0.032337 mol

                                                            = 0.0161685 mol

Theoretical mass of methanol formed =Theoretical moles x Molar mass

                                                            = 0.0161685 mol x (32.04216 g/mol)

                                                            = 0.5181 g

Hence, theoretical yield of methanol in given reaction = 0.5181 g

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