In this problem, assume that all gas volumes are measured at a temperature 320 K

Question

In this problem, assume that all gas volumes are measured at a temperature 320 K and pressure 120 kPa (i.e., the product(s) of the reaction as well as the reactants). The gases H_2 and O_2 react to form water vapor, H_2O. The chemical equation that describes this reaction is 2 H_2 + O_2 rightarrow 2 H_2O. This can be interpreted as follows: Two molecules of hydrogen (H_2) react with one oxygen (O_2) molecule to form two molecules of water (H_2O). Suppose that 600 ml of H_2 take part in a reaction with O_2. Find an expression for the number of moles in the hydrogen sample. How many hydrogen molecules are in the sample? How many moles of O_2 are needed to react with this amount of hydrogen? What is the volume of O_2, needed? (You should be able to find a numerical answer.) How many moles of H_2O are produced in this reaction? What is the volume of the H_2O produced? If the reaction were carried out again at a different (constant pressure and temperature, would the ratio of volumes of hydrogen and water change? Explain your reasoning. Do your answers to part C depend on the relative size or the relative mass of the reactant particles? (For example, would your answer change if the reaction were between N_2 and O_2 to produce nitrous oxide gas, N_2O? Explain your reasoning.

Explanation / Answer

Ans. Given,

            Pressure, P = 120 kPa = 1.18431 atm             ; [1 kPa = 0.00986923 atm]

            Temperature, T = 320 K

            Stoichiometry of reaction: 2H2 + O2 ---> 2 H2O

            It’s assumed that the gases behave as ideal gas.

Part C: Volume of H2 = 600 mL = 0.6 L                      ; [1L = 1000 mL]

Using ideal gas equation Ideal gas Law: pV = nRT          - equation 1

            Where, p = pressure in atm = 1.18431 atm

            V = volume in L             = 0.600 L

            n = number of moles = ?

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

            T = absolute temperature           = 320 K

Putting the values in equation 1

            (1.18431 atm) x (0.600 L) = n (0.0821 atm L mol-1K-1) (320K)

            Or, n = [(1.18431 atm) x (0.600 L)] / [(0.0821 atm L mol-1K-1) (320K)]

            Or, n = 0.710586 atm L / 26.272 atm L mol-1

            Hence, n = 0.027 mol

Thus, number of moles of H2 = 0.027 mol

According to the stoichiometry, 2 mol H2 produces 2 mol H2O;

Or,

Number of moles of H2O produced = Number of moles of H2

Hence, Number of moles of H2O produced = 0.027 mol

Volume of H2O produced, V = nRT/P

            Or, V = [(0.027 mol) (0.0821 atm L mol-1K-1) (320K)] / (1.18431 atm)

                        = 0.5989 L

                        = 600 mL [equal mole of two gases occupy same volume at identical P and T]

Note: equal moles of two different gases occupy same volume under given set of identical conditions of P and T. If number of moles were written in more decimal points, a value closer to 600 mL would have been attained.

Ans. D. No, equal moles of two different gases occupy same volume under given set of identical conditions of P and T. Since, moles of H2 and H2O are the same, they constitute the same volume under identical conditions.

Ans. E. No, the stoichiometry follows number of moles, but not the mass of reactants or products.

As long as the number of moles of N2 is equal to that of H2 and follows same stoichiometry, the result remains unaffected by the relative mass of reactants.

Related Questions

Navigate

• Browse (All)
• Browse I
• Subjects

---

---

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.