(Old) Chem and Ochem CR Ch 2 Prob 12

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jrashaad
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(Old) Chem and Ochem CR Ch 2 Prob 12

Postby jrashaad » Sat Dec 23, 2017 4:41 pm

12. In Reaction 2, if 54 g of water was formed, how much ethane and oxygen must have reacted?

In the solution it says that "From the equation, we can say that 1 mole of ethane must have reacted to form 3 moles of water. We can also say that 3.5 moles of oxygen must have reacted in the formation of 3 moles of water."

I don't really understand this problem and need help in calculating the # of moles of ethane and oxygen to yield the 3 moles of water.
NS_Tutor_Andrew
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Re: (Old) Chem and Ochem CR Ch 2 Prob 12

Postby NS_Tutor_Andrew » Tue Dec 26, 2017 12:42 pm

Hi jrashaad,

Thanks for the excellent question! Stoichiometry is definitely a topic where practice pays off.

The reaction we're given is 2 C2H6 + 7 O2 --> 4 CO2 + 6 H2O.

So, when we're told that 54 g of water was formed from this reaction, our first step is to convert to moles. The MW of water is 18, so we have 54/18 = 3 moles. Given this information, there are two ways we can proceed to identify how many moles of reactants there must have been: the short way and the long way.

Let's start with the short way. We can observe that 3 moles of H2O is exactly half of the stoichiometric constant we're given in the reaction (6 H2O). So we can just divide everything by 2 and get 1 C2H6 + 3.5 O2 --> 2 CO2 + 3 H2O, allowing us to recognize that one mole of ethane and 3.5 moles of O2 were used. The proportions are exactly the same as in the reaction we were given -- the only reason why the reaction was written as 2 C2H6 + 7 O2 --> 4 CO2 + 6 H2O instead of 1 C2H6 + 3.5 O2 --> 2 CO2 + 3 H2O is that we conventionally balance reactions so that they have the smallest possible whole-number coefficients, and 3.5 isn't a whole number.

The long way would be to use dimensional analysis. The balanced equation we're given tells us that 6 moles of H2O correspond to 7 moles of O2 or 2 moles of C2H6. So we could make the following calculations: 3 moles H2O * (2 moles C2H6 / 6 moles H2O) = 1 mole C2H6 and 3 moles H2O * (7 moles O2 / 6 moles H2O) = 3.5 moles O2. The advantage of doing it this way is that it could scale more easily to random numbers of moles -- the "short way" I described above is useful if you can clearly see how you have to multiply/divide the original reaction to correspond to the info in the Q, whereas you could use the "long way" even if you were given something initially bizarre like 314 moles of H2O.

Hope this is helpful!!
Andrew D.
Content Manager, Next Step Test Prep.

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