Question

W Yuestion 18. 20. These questions relate to combustion of hydrocarbons. a. LPG (liquid petroleum gas) is mostly propane, CzHg. Balance this equation. C3H2(g) + O2(g) → CO2(g) + H20(3) b. Cigarette lighters burn butane, C4H10. Write a balanced equation, assuming complete combustion, that is, plenty of oxygen. c. With a limited supply of oxygen, both propane and butane can burn incompletely to form carbon monoxide. Write balanced equations for both reactions.

Answer 1

(please post one question per post, since our guidelines and time don't allow to answer all questions. Moreover, question 22 is incomplete also.)

QUESTION 20:

(a):

Balanced equation:

CzHg 9) +502 (9) 3CO2(g) + 4H2O (9)

(b):

Balanced equation of combustion of butane (C₄H₁₀):

C4H10 (g) + 302 (9) — 4C02 (9) +5H20 (9)

OR (multiplying by 2 to get whole number stoichiometry)

2C4H10 (9) + 1302 (9) ► 8CO2(g) + 10 H2O (9)

(c):

Balanced equation for the partial combustion of propane:

C3H8 (9) +502 (9) ► 300 (9) + 4H20 (9)

OR (multiplying by 2 to get whole number stoichiometry)

2C3H8 9) +702 g) + 6CO(g) + 8 H2O (9)

Balanced equation of partial combustion of butane (C₄H₁₀):

C4H10 (9) + 02 (9) — 400 (g) + 5 H20 (9)

OR (multiplying by 2 to get whole number stoichiometry)

2C4H10 (9) + 902 g) → 8CO(g) + 10 H2O (9)

QUESTION 21:

(a)

C2H4 (9) + O2(g) → 2C s) + 2H2O (9)

(b)

C2H4 (9) +202 (9) ► 200 g) + 2 H2O (9)

(c)

C2H4 (9) + 302 (9) ► 2CO2(g) + 2 H2O (9)