Propane gas, $\ce{C3H8}$, burns in oxygen to produce carbon dioxide and water vapour as follows:

$$\ce{C3H8 (g) + 5O2 (g) = 3 CO2(g) + 4 H2O (g) + \pu{2200 kJ}}$$

If $\pu{1.5 mol}$ of propane is consumed in this reaction, how many moles of $\ce{H2O}$ are produced?

I'm completely stuck on this question, I have no idea how to start. Any help would be greatly appreciated.

  • $\begingroup$ Let's start with the balanced reaction: how many moles of water 1 mol of propane produces? You have $1~\ce{C3H8}$ on the left (reactants), and $4~\ce{H2O}$ on the right (products). Abstract from the energy, it's irrelevant here. $\endgroup$ – andselisk Dec 10 '17 at 10:27
  • $\begingroup$ Wouldn't that be 4 moles of water produced? $\endgroup$ – Grimestock Dec 10 '17 at 10:30
  • $\begingroup$ Yep, 4 it is. Now if there is 1.5 mol of propane (e.g. 1.5 times more), how many moles of water would it be? Remember that coefficients reflect molar ratios. $\endgroup$ – andselisk Dec 10 '17 at 10:32
  • $\begingroup$ 6 moles of water produced? $\endgroup$ – Grimestock Dec 10 '17 at 10:33
  • 1
    $\begingroup$ Wow. That was a lot simpler than I thought. Thank you! $\endgroup$ – Grimestock Dec 10 '17 at 10:36

The ratio between propane and water can be taken out of your reaction equation.

$$\frac{n(\ce{H20})}{n(\ce{C3H8})} = \frac{4}{1}$$

So if you burn 1.5 mole propane it will result in
$$\frac{4}{1} \cdot 1.5 = 6$$ mole of water being produced


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