Unhealthy exhaust results from the combustion of octane (C_9H_10) in automobile

Question

Unhealthy exhaust results from the combustion of octane (C_9H_10) in automobile engines via the following reaction. C_8H_18(1) + 25O_2(g) 16CO_2(g) + 18H_2O(g) Assuming the gasoline is pure octane and that a suburban dweller burns about 6.44 times 10^4 grams of octane (about 20 gallons) per week. How many grams of CO_2 are produced (per week) to warm the globed At 25 degree C and 1.00 atm, how many ml of CO_2 are produced? Give your answer in scientific notation. b. During combustion, N_2 and O_2 in the air react to make the nasty brown gas NO_2 which is smog. The smog rises up into the atmosphere, combines with water vapor and rains down as nitric acid (HNO_3) to acidify all the lakes and streams and kill the fish and plants. The following reactions are relevant. A. O_2(g) + N_2(g) 2NO(g) B. 2NO(g) + O_2(g) 2NO_2(g) C. NO_2(g) + H_2O(g) HNO_3(aq) + NO(g) Balance reaction C. Calculate the moles of HNO_3 that form if 2.2 times 10^4 moles of NO_2 rise up into the air and react completely with an unlimited amount of water vapor. c. A 200 mL sample of the lake water is collected titrated directly with 0.01M NaOH and it takes only 3 mL of the NaOH titrant to reach the equivalence point. Calculate the molarity (M) of HNO_3 in the lake water. d. Nitric acid is a strong acid and engages in the following 'proton' transfer (dissociation) reaction with water (in which it is dissolved). HNO_3(aq) + H_2O(1) H_3O^+(aq) + NO_3^-(aq) Suppose the concentration of HNO_3 in the lake water is 0.08 mg'mL. Calculate the molarity of H_5O^+ and the pH of the lake water.

Explanation / Answer

(a)

2 C8H18 (l) + 25 O2 (g) -------------> 16 CO2(g) + 18 H2O(g)

from the balanced equation provided,

2 mol C8H18 = 16 mol CO2

0r

228 g. of C8H18 = 704 g. of CO2

Then, 6.44 * 104 g. of C8H18 = 6.44 * 104 * 704 / 228 = 1.99 * 105 g. of CO2 per week.

Using ideal gas equation,

P V = n R T
P V = (m/M) R T

V = m R T / (M P )

V = 1.99 * 105 * 0.0821 * 298.15 / (44 * 1.00)

V = 1.11 * 105 L

V = 1.11 * 108 mL

Related Questions

Navigate

• Browse (All)
• Browse U
• Subjects

---

---

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