1) Butane, C4H10, is a component of natural gas that is used as fuel for cigaret

Question

1) Butane, C4H10, is a component of natural gas that is used as fuel for cigarette lighters. The balanced equation of the complete combustion of butane is

2C4H10(g)+13O2(g)8CO2(g)+10H2O(l)

At 1.00 atm and 23 C, what is the volume of carbon dioxide formed by the combustion of 2.60 g of butane?

2)How many air molecules are in a 10.0×12.0×10.0 ftroom? Assume atmospheric pressure of 1.00 atm, a room temperature of 20.0 C, and ideal behavior.

Volume conversion:There are 28.2 liters in one cubic foot.

Explanation / Answer

Given :

2C4H10(g)+13O2(g)--- >8CO2(g)+10H2O(l)

4 asdfsa Pressure = 1.00 atm

T = 23.0 deg C = 237.15 + 23.0 deg C = 296.15 K

Mass of butane = 2.60 g

Calculation of moles of butane = 2.60 g / molar mass of butane

= 2.60 g x 1 mol / 58.12 g

= 0.04474 mol

Calculation of moles of CO2

We use mol ratio to get moles of CO2 from moles of butane

Mol of CO2 = 0.04474 mol x 8 mol CO2 / 2 mol butane

= 0.1789 mol CO2

Now we use ideal gas law to get moles of CO2

pV = nRT

here p is pressure in atm, V is volume in L, T is Temperature in K , n is number of moles

R = 0.08206 L atm per (Kmol)

V = nRT / p

=[ 0.1789 x 0.08206 x 296.15 / 1.00 ] L

= 4.35 L Co2

Volume of CO2 produced = 4.35 L

Q. 2

Given

Volume of room = (10.0×12.0×10.0 )ft3 room

1 ft3 = 28.32 L

Conversion of volume to L

Volume in L = (10.0×12.0×10.0 )ft3 x 28.2 L / 1 ft3

= 33840 L

T = 20.0 deg C = 237.15 + 20.0 deg C = 293.15 K

P = 1.00 atm

Lets use ideal gas law to find out moles of air molecules.

n = pV / RT

= [1.00 x 33840 / (0.0806 x 293.15)] mol

= 1406.72415 mol

Number of molecules of air

= 1406.72415 mol x 6.02E23 air molecules / 1 mol

=8.47 E 26

Number of air molecule in room = 8.47 x 1026 molecules.

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