For Question 1
First, write the balanced stoichiometric equation for the combustion.
For complete combustion of hydrocarbons such as propane (C3H8), oxygen (O2) reacts with the hydrocarbon to produce carbon dioxide (CO2) and water (H2O).
C3H8 + O2 → CO2 + H2O (unbalanced)
The balanced equation is
C3H8 + 5O2 → 3CO2 + 4H2O
From the stoichiometric coefficients of this equation, we see that 1 mole of propane (C3H8) yields 3 moles of carbon dioxide. That is the mole ratio of
C3H8: CO2 is 1:3
For the remainder of this question, you will need the molar masses of C, H and O.
From the Periodic Table of the elements, these are:
C - 12.01g
H - 1.008 g
O - 16.00 g
Now, convert the 5g of propane to moles. For this, we need its molar mass.
Molar mass of C3H8 = 3(12.01g) + 8(1.008g) = 44.09 g
g C 3 H 8 x 1mole C3H8 = 0.1134 moles C3H8
g C 3 H 8
Now use the mole ratio of C3H8 : CO2 to determine the number of moles of CO2 produced from the 0.1134 moles C3H8.
Moles of CO2 produced = 0.1134
moles C 3 H 8 x 3 moles CO2 = 0.340 moles CO2
moles C 3 H 8
Answer = 0.340 moles CO2
Alternatively, you can combine the two steps above as follows:
g C 3 H 8 x 1 mole C 3 H 8 x 3 moles CO2 = 0.340 moles CO2
g C 3 H 8 1 moles C 3 H 8
To find the number of grams of CO2 produced, convert 0.340 moles CO2 to grams
Molar mass of CO2 = 12.01g + 2 (16.00 g) = 44.01 g
moles CO 2 x 44.01 g = 14.97 g CO2
mole CO 2
Answer = 14.97 g CO2