Question

Consider the following reaction representing the combustion of propane: ????????3HH8 + O2 → CO2 + H2O a. Balance the equation b. How many moles of oxygen are required to burn 1 mol of propane? c. How many grams of oxygen are required to burn 100 g of propane? d. At standard temperature and pressure, what volume of oxygen would be required to burn 100 g of propane? If air is 21 percent oxygen, what volume of air at STP would be required? e. At STP, what volume of carbon dioxide would be produced when 100 g of propane are burned? f. The standard enthalpy of propane is -103.8 kJ/mol. Find the gross heat released when 1 kg is burned.

Answer 1

Answer:

propane C3H8 +O2 =CO2 +H2O

Explanation:

C3H8 + 5O2 = 3CO2 + 4H2O

3C & 10 O  & 8 H on each side

Answer 2

Answer:

a) C₃H₈ + 5 O₂ → 3 CO₂ + 4 H₂O

b) You need 5 moles of O₂ per mole of propane.

c) 181 g of O₂

d) 254 L of O₂ and 1210 L of air

e) 152 L of carbon dioxide

f) The gross heat released is 2354 kJ

Explanation:

C₃H₈ + O₂ → CO₂ + H₂O

a) To balance a combustion reaction you must add CO₂ as carbons of hydrocarbons you have. Then, you should add waters as half of hydrogens of the hydrocarbon and, in the last, balance oxygen with O₂, thus:

C₃H₈ + O₂ → 3 CO₂ + 4 H₂O

10 oxygens, so you sholud add 5 O₂:

C₃H₈ + 5 O₂ → 3 CO₂ + 4 H₂O

b) The equation balanced says that you need 5 moles of O₂ per mole of propane.

c) To burn 100g of propane you need:

100 g C₃H₈×$\frac{1 mol}{44,1g}$×$\frac{5 molO_2}{1mol C_{3}H_{8}}$×$\frac{16g}{1mol O_2}$= 181 g of O₂

d) 181g of O₂ are 11,34 moles. The volume you require is:

V =nRT/P

where:

n are moles (11,34 moles)

R is gas constant (0,082 atmL/molK)

T is temperature (273 K at STP)

P is pressure (1 atm at STP)

V is 254 L of oxygen.

The liters of air are:

254L O₂ ₓ $\frac{100 air}{21 O_2}$ = 1210 L of air

e) The volume of CO₂ produced is:

100 g C₃H₈×$\frac{1 mol}{44,1g}$×$\frac{3 molCO_2}{1mol C_{3}H_{8}}$= 6,80 moles of CO₂

V =nRT/P

where:

n are moles (6,80 moles)

R is gas constant (0,082 atmL/molK)

T is temperature (273 K at STP)

P is pressure (1 atm at STP)

V is 152 L of carbon dioxide.

f) C₃H₈ + 5 O₂ → 3 CO₂ + 4 H₂O ΔH = -103,8 kJ/mol

1 kg of C₃H₈ are:

1000 g × $\frac{1mol}{44,1 g}$ = 22,68 moles

Thus, the gross heat released is:

103,8 kJ/mol × 22,68 moles = 2354 kJ

I hope it helps!