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.
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.
1) Butane, C4H10, is a component of natural gas that is used as fuel for cigarette...
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 3.40 g of butane?
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