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A 9.725-g gaseous mixture contains ethane () and propane (). Complete combustion to form carbon dioxide and water requires 1.115 moles of oxygen gas. Calculate the mass percent of ethane in the original mixture.

Sagot :

Answer:

% = 33.83%

Explanation:

To do a better understanding of this, we can treat the mixture of the combustion as two separate reactions, in that way, we can have an idea of what is happening and how to calculate the mass percent.

So the combustion reactions in this mixture are:

2C₂H₆ + 7O₂ ---------> 4CO₂ + 6H₂O

C₃H₈ + 5O₂ ----------> 3CO₂ + 4H₂O

Now that we have both reactions (And balanced) we can hace an idea of the mole ratio between every compound in the mix.

For practical purposes, let's call "a" the mass of ethane, and "b" the mass of propane. The innitial mix have 9.725 g, so this means that:

a + b = 9.725 g    (1)

Now that we have this, we can write a relation between the moles of oxygen and the moles of the gases. If we have 1.115 moles of oxygen, and also know the mole ratio of oxygen to "a" and "b", so:

moles O₂ = moles a (moles O₂/moles a) + moles b (moles O₂/moles b)   (2)

And we know that moles a and moles b are:

moles a = a / MW

moles b = b / MW

The MW of a is 30 g/mol and the MW of b is 44 g/mol

Replacing the given data we have:

1.115 moles O₂ = (a/30)(7 moles O₂/2 moles a) + (b/44)(5 moles O₂/1 mole b)

1.115 moles O₂ = (0.1167a) moles O₂ + (0.1136b) moles O₂

To keep solving this, we can use expression (1) to solve for b, and then, replace here and have only one equation with 1 incognite:

a + b = 9.725 g  

b = 9.725 - a   (3)

Replacing above we have:

1.115 = 0.1167a + 0.1136(9.725 - a)

1.115 = 0.1167a + (1.1048 - 0.1136a)

1.115 - 1.1048 = 0.1167a - 0.1136a

0.0102 = 0.0031a

a = 3.29 g

Now, that we have the mass of the ethane, we can calculate the mass percent:

% = (3.29 / 9.725) * 100

% = 33.83%

Hope this helps

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