How to calculate the amount of water produced by the incomplete combustion of propane? I was wondering if someone could walk me through how to do the following two questions?


  
  The complete combustion of pentane has a enthalpy change of $-3509~\mathrm{mol^{-1}}$. Write an equation for the complete combustion of pentane. Write a possible equation for the imcomplete combustion of pentane.
  


For this question, would the following equation suffice:
$$\ce{ C5H12 (g) + 8O2(g) -&gt; 5CO2 (g) + 6H2O (g)  }$$

For an incomplete reaction, the products will be water, carbon/and or carbon monoxide due to insufficient oxygen, so would this equation be correct:
$$\ce{C5H12 (g) + 11/2 O2 (g) -&gt; 5CO (g) + 6H2O (g)}$$


  
  $17~\mathrm{L}$ of Propane are incompletely combusted in $34~\mathrm{L}$ of oxygen at $25~^\circ\mathrm{C}$ and $100~\mathrm{kPa}$ into only water and carbon. Write a balanced chemical question to represent this including state symbols. What is the mass of the water produced? What observable change would indicate that a chemical change occurred?
  


Would an equation like this work: $$\ce{C3H8 (g) + 7O2 (g) -&gt; 3CO2 (g) + 4H2O (g)}$$

The ratio for the above equation will be 1:7:3:4.

I am given the volume of the propane to be $17~\mathrm{L}$ and the volume of the oxygen to be $34~\mathrm{L}$. Since the question is a molar calculation, I think I have to convert the $~\mathrm{L}$ into the SI units of grams. This means $17~\mathrm{L} \Rightarrow 17000~\mathrm{g}$ and $34~\mathrm{L} \Rightarrow 34000~\mathrm{g}$.

The I have to calculate the number of moles for propane and oxygen (I think).

$$
\text{number of moles} = \frac{\text{mass}}{\text{molar mass}}\\
\text{number of moles of propane} = \frac{17000}{44.1} = 385.4~\mathrm{\frac{mol}{gram}}\\
\text{number of moles of oxygen} = \frac{34000}{16} = 2125~\mathrm{\frac{mol}{gram}}$$

Finally I have to find the mass of the resulting water, which is the part I am stuck at. I know I have to find the number of moles times the molar mass of water to find the total mass of the water, but I am just somewhat stumped as to how to find the number of moles of water given what I have worked out above.

For chemical changes, I think that bubbles of gas may appear, a temperature change will occur since this is a combustion reaction, and maybe a change of mass/volume?
Yes the complete combustion equation is correct as you stated and the equation of the incomplete combustion is certainly one possibility. A more general formula is:
$$\begin{multline}\ce{C5H12 + $\left(\frac{x}{2}+y+3\right)$\,O2 -&gt; $(5-x-y)$\,C + $x$\,CO +$y$\,CO2 + 6H2O}\\
\left\{x,y\in\mathbb{R}^{+}_0; (x+y)\leq5\right\}
\end{multline}$$
This is assuming all the hydrogen will be converted into water. In some cases you might have incomplete combustion, where ethene or ethyne (ar many others) are formed, but I think this is taking it a bit far at this point.
The above case reduces to your formula for $x=5$ and $y=0$.



For the second question, your equation has a small error, correct is:
$$\ce{C3H8 (g) + 5O2 (g) -&gt; 3CO2 (g) + 4H2O (g)}$$
But this is not the equation you need, as the question explicitly states, that the combustion products are carbon and water. So you working equation becomes:
$$\ce{C3H8 (g) + 2O2 (g) -&gt; C (s) + 4H2O (g)}$$
So you need twice the number of moles of oxygen than propane, to produce four times as many moles of water than you had propane.

You cannot convert a volume of gas into a mass by an easy conversion factor, you have to use the ideal gas law:
$$pV =n\mathcal{R}T$$
You can then use $m =n\cdot M$ to convert to grams, but you don't yet have to do that.

You have been given the temperature $T=25~^\circ\mathrm{C}=298.15~\mathrm{K}$ and pressure $p=100~\mathrm{kPa}$ and the volumes for the reactants, so you can calculate the amount of substance for propane and oxygen.


   $n(\ce{C3H8})\approx 0.7~\mathrm{mol}$ and $n(\ce{O2})\approx 1.4~\mathrm{mol}$


From the reaction equation you can now determine the amount of substance of water, that will be produced.


   $n(\ce{H2O})\approx 2.8~\mathrm{mol}$


You can convert that easily into a mass now.


   $m(\ce{H2O})\approx 50~\mathrm{mol}$


There are two notable changes you will observe. Since you are combusting a gas, obviously no bubbles can appear. If you do the combustion in a see-through container you will see black stains due to the solid carbon that is produced by the reaction and liquid water drops will form on colder surfaces.
You are of course right, that you will also have a rise in temperature.

Note: The here given values are crude approximations only, that might not satisfy the task at hand. You need to do the calculation yourself to find the correct answer.