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bogdanovich [222]
3 years ago
9

Using the equation, C5H12 + 8O2 Imported Asset 5CO2 + 6H2O, if an excess of pentane (C5H12) were supplied, but only 4 moles of o

xygen were available, how many moles of water would be produced?
Chemistry
2 answers:
Irina18 [472]3 years ago
8 0
<span>the balanced equation for the combustion of pentane is as follows:

C5H12 + 8O2 --> 5CO2 + 6H2O

excess pentane is allowed to react with 4 moles of oxygen. this means that pentane is the excess reactant and O</span>₂<span> is the limiting reactant. Limiting reactant is the reagent that is fully used up in the reaction and amount of product formed depends on the amount of limiting reactant present.
Stoichiometry of O</span>₂<span> to H</span>₂<span>O is 8: 6.
If 8 mol of O</span>₂<span> forms 6 mol of H</span>₂<span>O then 4 mol of O</span>₂<span> forms - 6/8 x 4 = 3 mol of H</span>₂<span>O
therefore 3 mol of H</span>₂<span>O is formed</span>
ser-zykov [4K]3 years ago
7 0
Answer: 3 <span>moles of water would be produced in present case.
</span>
Reason:
Reaction involved in present case is:
<span>                            C5H12 + 8O2 </span>→<span> 5CO2 + 6H2O

In above reaction, 1 mole of C5H12 reacts with 8 moles of oxygen to give 6 moles of water.

Thus, 4 moles of oxygen will react with 0.5 mole of C5H12, to generate 3 moles of H2O.</span>
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                          H_{1/2} = \frac{0.693}{Rate Constant }

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                         H_{1/2} = \frac{0.693}{9.45*10^{-5}}

                                 H_{1/2}= 7333.3sec

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a) Whatis the composition in mole fractions of a solution of benzene and toluene that has a vapor pressure of 35 torr at 20 °C?
iogann1982 [59]

Answer:

molar composition for liquid

xb= 0.24

xt=0.76

molar composition for vapor

yb=0.51

yt=0.49

Explanation:

For an ideal solution we can use the Raoult law.

Raoult law: in an ideal liquid solution, the vapor pressure for every component in the solution (partial pressure) is equal to the vapor pressure of every pure component multiple by its molar fraction.  

For toluene and benzene would be:

P_{B}=x_{B}*P_{B}^{o}

P_{T}=x_{T}*P_{T}^{o}

Where:

P_{B} is partial pressure for benzene in the liquid  

x_{B} is benzene molar fraction in the liquid  

P_{B}^{o} vapor pressure for pure benzene.  

The total pressure in the solution is:

P= P_{T}+ P_{B}

And  

1=x_{B}+x_{T}

Working on the equation for total pressure we have:

P=x_{B}*P_{B}^{o} + x_{T}*P_{T}^{o}

Since x_{T}=1-x_{B}

P=x_{B}*P_{B}^{o} + (1-x_{B})*P_{T}^{o}

We know P and both vapor pressures so we can clear x_{B} from the equation.

x_{B}=\frac{P- P_{T}^{o}}{ P_{B}^{o} - P_{T}^{o}}

x_{B}=\frac{35- 22}{75-22} = 0.24

So  

x_{T}=1-0.24 = 0.76

To get the mole fraction for the vapor we know that in the equilibrium:

P_{B}=y_{B}*P

y_{T}=1-y_{B}

So  

y_{B} =\frac{P_{B}}{P}=\frac{ x_{B}*P_{B}^{o}}{P}

y_{B}=\frac{0.24*75}{35}=0.51

y_{T}=1-0.51=0.49

Something that we can see in these compositions is that the liquid is richer in the less volatile compound (toluene) and the vapor in the more volatile compound (benzene). If we take away this vapor from the solution, the solution is going to reach a new state of equilibrium, where more vapor will be produced. This vapor will have a higher molar fraction of the more volatile compound. If we do this a lot of times, we can get a vapor that is almost pure in the more volatile compound. This is principle used in the fractional distillation.

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3 years ago
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