Flow rate of the fuel consisting of ethanol and water burning in air of a given flow rate

I hope that someone could give me some hints or tips on how to approach this question, even though these types of questions are usually closed off as homework.

A burner is run by a fuel mixture of $$\pu{90 mol.\%}$$ ethanol and $$\pu{10 mol.\%}$$ water. Assume a complete combustion with air ($$\pu{79 mol.\%}$$ $$\ce{N2}$$, $$\pu{21 mol.\%}$$ $$\ce{O2}$$). The whole process takes place at $$\pu{100 kPa}$$.

Calculate the inlet flow rate of the fuel given that the flow rate of air is $$\pu{100 mol s-1}$$ and that the excess of air is $$20\%$$.

The answer key only gives the answer $$\pu{6.48 mol s-1}$$ of the fuel without explaining the calculations.

My attempt at the question was:

$$\ce{C2H5OH + H2O + 3O2 -> 2CO2 + 4H2O}$$

Knowing that we have $$20\%$$ excess in air, we could calculate the theoretical moles needed.

$${\text{fractional excess} = (n_\mathrm{feed} - n_\mathrm{theo})/n_\mathrm{theo}}$$

$$\to n_\mathrm{theo} = \pu{83.3 mol}$$

I then used the molar ratio from the reaction to calculate how much ethanol and water was needed. I just don't see how the answer becomes $$\pu{6.48 mol s-1}$$. Any help is greatly appreciated!

• A burner combusts fuel. What takes part in such reactions? Jan 11, 2019 at 14:54
• Also, the question looks like just stoichiometry. Why did you include a heat exchanger etc? Jan 11, 2019 at 15:01
• @EashaanGodbole I added the water in the reaction since they stated in the problem that the fuel consisted of both ethanol and water, although I know that a combustion reaction does not include water in the reactants. The diagram of the system was given, there are other questions which involve the heat exchanger specifically. Jan 11, 2019 at 15:16
• The inlet water should not be included in the balanced chemical reaction formula, and you balanced the formula incorrectly. Your first step should be to get that right. Jan 11, 2019 at 17:12

I think you overcomplicate the problem. There are many variables you don't need and many components that are duds (nitrogen, water, pressure). What important is that we burn ethanol:

$$\ce{C2H5OH + 3 O2 -> 2 CO2 + 3 H2O}$$

so that the ratio between ethanol and oxygen is $$1:3$$. I have taken chemistry engineering classes long time ago and I don't remember exact notations, so I'd rather denote molar flow rate with an n-dot notation:

$$\dot{n} = \frac{\mathrm{d}n}{\mathrm{d}t}$$

where $$n$$ – amount; $$t$$ – time. Since fuel is only 90% alcohol, then

$$\dot{n}(\text{fuel}) = \frac{\dot{n}(\ce{C2H5OH})}{x(\ce{C2H5OH})}$$

where $$x$$ denotes molar fraction. From the combustion reaction

$$\dot{n}(\ce{C2H5OH}) = \frac{\dot{n}(\ce{O2})}{3}$$

and for the air the following is fulfilled:

$$\dot{n}(\ce{O2}) = x(\ce{O2})\frac{\dot{n}(\text{air})}{1 + α}$$

where $$α$$ is the partial excess of air. Now we can put everything together and plug in the numbers:

\begin{align} \dot{n}(\text{fuel}) &= \frac{x(\ce{O2})\dot{n}(\text{air})}{3x(\ce{C2H5OH})(1 + α)}\\ &= \frac{0.21\cdot\pu{100 mol s-1}}{3\cdot 0.9\cdot (1 + 0.2)}\\ &= \pu{6.48 mol s-1} \end{align}

One hell of a burner.

You calculated 83.3 moles of air reacting with the ethanol. Since air is 21% oxygen, this means that 83.3 x 0.21 = 17.5 moles of oxygen react with the ethanol. Since each mole of ethanol reacts with 3 moles of oxygen, the number of moles of ethanol that react is 17.5/3=5.833 moles ethanol. Since the fuel mixture is 90% ethanol, there are 10 moles of fuel mixture for every 9 moles of ethanol. So the number of moles of fuel mixture are 5.833 x 10/9=6.48 moles of fuel mixture.