Note there are 2 questions: a & b. please show all steps as am studying for a qu

Question

Note there are 2 questions: a & b.

please show all steps as am studying for a quiz. Thank You

Consider the combustion of ethane (C2H6) in air to form CO2 and H2O. How much O2(g) (in grams) is used per gram of ethane? From this, what is the mass-based air:fuel ratio? What is the concentration of H2O (%, volume) in the exhaust if a stoichiometric amount of dry air (assume 21% O2 and 79% N2, by volume) is used? Compare this to the typical water vapor concentration of about 10,000 ppm (as you know from living in Atlanta, the water vapor concentration is highly variable).

Explanation / Answer

Ethane burns in oxygen to produce carbon dioxide and water.

First we want to balance the equation

__C2H6 + __O2   -->   __CO2 + __H2O

Oxygen is the simplest, so we will find its coefficient last

We need two CO2 s to take care of the two Cs in ethane.

We need three H2Os to take care of the six Hs in ethane

This leaves us with an odd number of oxygens, so we could have a fractional coefficient

C2H6 + 2.5 O2   ---> 2 CO2 + 3 H2O

note that there is one C2H6, but we don't usually write coefficients of one.

If we don't like fractions, we could double everything

2 C2H6 + 5 O2 ---> 4 CO2 +6 H2O

So we need five moles of oxygen for every two moles of ethane. That is the beautiful thing about moles, simple whole numebr ratios. Unfortunately, the nasty question asks for grams.

C =12 g/mol

H = 1 g/mol

O = 16 g/mol

C2H6 is 30g/mole

O2 is 32 g/mole

C2H6 g :O2 g = 2*30:5:32

so for every 160g O2 we need 60g C2H6

60/60 = 1, 160/60 = 2.67

so for every 1g ethane we need 2.67 g O2

mass based air:fuel ratio = 2.67:1   or we could use our original 160:60 = 2.67

b) For ease, let's say we burned 5 moles O2. We would then have produced 4 mole CO2 and 6 mole H2O, plus we have

more N2

The moles of gas we send through the system is X (equal moles take up equal volumes)

0.21X = 5

X = 5/0.21

X = 23.80

N2 moles = 23.8*0.79 = 18.8

So the exhaust is 6/total water

or 6/(6+4+18.8) = 20.8 %H20

20.8/100 water

is 208000/1000000

so 208,000 PPM

so water concentration is about 20 times more

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