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bogdanovich [222]
4 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]4 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]4 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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When the pressure that a gas exerts
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<h2>Hello!</h2>

The answer is:

When the pressure that a gas exerts  on a sealed container changes from

22.5 psi to 19.86 psi, the  temperature changes from 110°C to

65.9°C.

<h2>Why?</h2>

To calculate which is the last pressure, we need to use Gay-Lussac's law.

The Gay-Lussac's Law states that when the volume is kept constant, the temperature (absolute temperature) and the pressure are proportional.

The Gay-Lussac's equation states that:

\frac{P_1}{T_1}=\frac{P_2}{T_2}

We are given the following information:

We need to remember that since the temperatures are given in Celsius degrees, we need to convert it to Kelvin (absolute temperature) before use the equation, so:

P_1=22.5Psi\\T_1=110\°C=110\°C+273.15=383.15K\\T_1=65.9\°C=65\°C+273.15=338.15K

Now, calculating we have:

\frac{P_1}{T_1}*(T_2)=P_2\\\\P_2=\frac{P_1}{T_1}*(T_2)=\frac{22.5Psi}{383.15}*338.15=19.86Psi

Hence, the final pressure is equal to 19.86 Psi.

Have a nice day!

8 0
3 years ago
Read 2 more answers
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