(3 of 32) A student performed the experiment described in this chapter. A 3.50 m

Question

(3 of 32) A student performed the experiment described in this chapter. A 3.50 mL aliquot of 3.00% wt/wt solution of hydrogen peroxide (MM-34.016 g/mol) with density 1.10 g/mL. The water temperature was 26.0 °C after the reaction and the barometric pressure in the lab was 29.92 in. of Hg. Calculate the volume of O2 that should be generated (See Table IV.3 for selected water vapor pressures is available in Appendix IV of your lab manual). R= 0.08205 L-atm/mol-K 8.62 mL 17.2 mL 7.49 mL 15.0 mL 43.1 mL 3.75 mL 56.7 mL 4.93 mL 86.2 mL 172 mL 4.31 mL

Explanation / Answer

Mass of 3.50 mL of solution = density * volume = 1.10 * 3.50 = 3.85 g.

Then,

Mass of H2O2 = 3.85 * 3.00 / 100 = 0.116 g.

Moles of H2O2 = mass / molar mass = 0.116 / 34.0 = 0.00340 mol

2 H2O2 (l) -----------> 2 H2O (l) + O2 (g)

From the above balanced equation,

2 mol of H2O2 gives 1 mol of O2

then, 0.00340 mol of H2O2 gives 1 * 0.00340 / 2 = 0.00170 mol of O2

Pressure of O2 = barometric pressure - water vapor pressure at 26.00C

P = (29.92 * 25.4) - 25.5 = 734.5 mmHg = 734.5 / 760 atm = 0.966 atm

R = 0.08205 L.atm.K-1.mol-1

T = 26.0 + 273.15 = 299.15 K

Ideal gas equation,

P V = n R T
V = 0.00170 * 0.08205 * 299.15 / 0.966

V = 0.0432 L

V = 43.2 mL

So, the answer is (e)

Related Questions

Navigate

• Browse (All)
• Browse #
• Subjects

---

---

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