1) Propane (C3H8)is a minor component of natural gas and is used domestic cookin

Question

1) Propane (C3H8)is a minor component of natural gas and is used domestic cooking heating.

(A) Balance the following equation representing the combustion of propane in air:

C3H8+O2--->CO2+H2O

(B) How many grams of carbon dioxide can be produced by burning 3.95 moles of propane? (assume that oxygen is the excess reactant in this reaction)

(g)?

2) Calculate the mass in grams of iodine (I2) that will react completely with 23.7g of aluminum (Al) to form aluminum iodine (AlI3)

g I2

3) Consider the combustion of carbon monoxide (CO) in oxygen gas

    2CO(g)+O2(g)--->2CO2(g)

starting with 3.35 moles of CO, calculate the number of moles of CO2 produced if there is enough oxygen gas to react with all the CO

mol CO2?

4) Perchloroethylene, also known as perc, is the solvent used in most dry cleaning. It consists of C and Cl. Determine the empirical formula of perchloroethylene given that a 2.58g sample yields 1.37g CO2 in a combustion analysis experiment. Add to complete the formula (CCl)

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Explanation / Answer

1

(A) Balanced equation

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

(B) Convert from 3.95 moles C3H8 into grams CO2.

(3.95 moles C3H8)(3 moles CO2 / 1 mole C3H8)(44g CO2 / mole CO2) = 480grams CO2

2.

2Al + 3I2 ----> 2AlI3. You balanced it correctly.

We first have to find the number of moles of Al reacting
No of moles = mass / Molar mass
= 23.7 / 26.98
=0.88 moles

2Al + 3I2 ----> 2AlI3
2 : 3 : 2
2(0.88) of Al reacts with 3(0.88) moles of Iodine according to the equation.
Moles of Iodine = 3 X 0.88 = 2.64moles

Mass of Iodine = moles X molar mass
= 2.64 X 126.9
= 380g

3.

the ratio between CO and CO2 is 2 : 2 ( or 1 : 1)

3.35 moles CO + 1.80 moles O2 >> 3.35 moles CO2

4.

mass of C in 1.37g CO2 = 12/44*1.37 = 0.3736g
Mass of Cl= 2.58-0.3736 = 2.206

Divide each by respective atomic mass
C = 0.3736/12 = 0.0311
Cl = 2.206 / 35.45 = 0.0622

divide by smallest:
C = 1
Cl = 2

Empirical formula = CCl2

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