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lesantik [10]
3 years ago
10

How many moles of ethylene (C2H4) can react with 12.9 liters of oxygen gas at 1.2 atmospheres and 297 Kelvin? C2H4(g) + 3O2(g) y

ields 2CO2(g) + 2H2O(g) 0.21 moles 0.32 moles 0.63 moles 0.84 moles
Chemistry
1 answer:
7nadin3 [17]3 years ago
5 0
1) Convert 12.9 liters of Oxygen to mol at the given conditions:

PV = nRT ⇒ n = PV/RT

n = [1.2atm*12.9 l] / [0.082 atm l /K mol * 297K]

n = 0.636 mol of O2

2) use the stoichiometry derived from the balanced chemical equation

 1mol C2H4 / 3  mol O2 =  x mol C2H4 / 0.636 mol O2

x = 0.636 / 3 mol O2 = 0.212 mol O2.

Answer: 0.212 mol O2
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Explanation:

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Volume of solution = 1 L

Initial concentration of Cl_2 = \frac{0.14mol}{1L}=0.14M

The given balanced equilibrium reaction is,

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Initial conc.           0.14 M           0 M       0M    

At eqm. conc.     (0.14-x) M        (x) M        (x) M

The expression for equilibrium constant for this reaction will be,

K_c=\frac{[CO]\times [Cl_2]}{[COCl_2]}

Now put all the given values in this expression, we get :

4.63\times 10^{-3}=\frac{x)^2}{(0.14-x)}

By solving the term 'x', we get :

x = 0.023 M

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Explanation:

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Answer:

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