Question

Propane C3H8 is a hydrocarbon tht is commonly used as a fuel
(a)write a balanced equation for the complete combustion used Propane gas.
(b) Calculate the volume of air at 25C and 1.00 atmosphere that is needed to completely combust 25.0 grams of propna. Assume that air is 21.0 percnt O2 by volume. (Hint: we will see how to do this calcullation in a later chapter on gases -- for now use the infomation that 1.00L of air at 25C and 1.00atm contains 0.275 g of O2 pe liter.)
(c) The heat of combustion of propane is -2,219.2 kJ/mol. Calculate the heat of formation, Hof of propane given that hof of H2O = -285.8 kJ/mol and hof of CO2 = -393.5 kj/mol
(d) Assuming that all of the het releasd in burning 25.0 grams of propane is transferred to 4.00 kilograms of water calculate the increase in temperature of water .

Answer 1

Answer:C3H8 +5O2-- >3CO2+4H2O, 330L of air,Hof C3H8= 2323.7 KJ/mol,dT=75.30

Explanation:

First We need to find a balanced equation to depict the combustion of the propane:

C3H8 +5O2-- >3CO2+4H2O

b)

$25g C3H8*\frac{1mol C3H8}{44gC3H8} *\frac{5molO2}{1molC3H8}* \frac{32gO2}{1mol O2}*\frac{1L air}{0.275g}$

330L of air

25 g C3H8 are equal to 44g and 5 mol of O2 will react with a mol of C3H8, each mol of O2 weigh 32g and for each L of air contains 0.275g of O2

c) $Hof C3H8= -(-285.8\frac{KJ}{mol}*4mol +-393.5*3mol$

the enthalpy of formation of propane will be the negative of the sum of the enthalpy of formation multiplied by the mols of the H2O and CO2

Hof C3H8= 2323.7 KJ/mol

d) We know that each 44g of C3H8(1mol) transfer 2219.2 KJ then:

$25g\frac{2219.2KJ}{44g C3H8}$ = 1260.9KJ

We use the equation of heat transfered

Q=mCpdT

1260.9KJ= 4KJ *4.186KJ/Kg*°C  *dT

We clear the equation

Q/mCp=dT

dT=75.30

Answer 2

The balanced equation for the complete combustion of Propane gas is; C₃H₈ + 5O₂ → 3CO₂ + 4H₂O

What is the balanced chemical equation?

A) The balanced equation for the complete combustion of Propane gas is; C₃H₈ + 5O₂ → 3CO₂ + 4H₂O

B) 25 g C₃H₈ * 1 mol C₃H₈/44g * 5 mol O₂/1 mol C₃H₈ = 2.84 mol O₂

Volume of O₂ = nRT/P

Volume of O₂= (2.84 mol * 0.0821 L.atm/mol.K * 288 K)/1 atm

Volume of O₂ = 67.15 L

Since oxygen is 21% of air, then;

Volume of air = 67.15/0.21

Volume of air = 32O L

C) We want to find the heat of combustion;

-2219.2 = [3(-393.5) + 4(-285.8)] - [X + 0]

-2219.2 = -1180.5 - 1143.2 - X

-2219.2 = -2,323.7 - X

X = -104.5 kJ/mol = ΔH°_comb

D) Let us first find the quantity of heat;

q = 25 g C₃H₈ * 1 mol/44g * 2,219.2 kJ/1 mol

q = 1260.91 kJ

formula for quantity of heat is;

q = m*c*Δt

where;

m is mass

c is specific heat capacity

Δt is change in temperature

Thus;

1260.91 = 4 * 4.184 * Δt

Δt = 1260.91/(4 * 4.184)

Δt = 75.34°

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