Question #40f45

1 Answer
Jul 15, 2016

"96 L O"_2

Explanation:

The first thing to do here is write a balanced chemical equation that describes the combustion of ethanol, "CH"_3"CH"_2"OH"

"CH"_ 3"CH"_ 2"OH"_ ((l)) + color(red)(3)"O"_ (2(g)) -> 2"CO"_ (2(g)) + 3"H"_ 2"O"_ ((l))

As you can see, the reaction consumes color(red)(3) moles of oxygen gas for every mole of ethanol that undergoes combustion.

Use the molar mass of ethanol to convert the sample from grams to moles

65 color(red)(cancel(color(black)("g"))) * ("1 mole CH"_3"CH"_2"OH")/(46.07color(red)(cancel(color(black)("g")))) = "1.411 moles CH"_3"CH"_2"OH"

According to the aforementioned mole ratio, this many moles of ethanol would require

1.411 color(red)(cancel(color(black)("moles CH"_3"CH"_2"OH"))) * (color(red)(3)color(white)(a)"moles O"_2)/(1color(red)(cancel(color(black)("mole CH"_3"CH"_2"OH")))) = "4.233 moles O"_2

Now, STP conditions are usually given to students as a pressure of "1 atm" and a temperature of 0^@"C". This definition gets a molar volume of a gas at STP equal to "22.4 L".

However, the current definition of STP conditions works with a pressure of "100 kPa" and a temperature of 0^@"C". Under these conditions, the molar volume of a gas is equal to "22.7 L".

I'll use the current definition, but feel free to redo the calculation using the old one if that's what was given to you

4.233 color(red)(cancel(color(black)("moles O"_2))) * "22.7 L"/(1color(red)(cancel(color(black)("mole O"_2)))) = color(green)(|bar(ul(color(white)(a/a)color(black)("96 L")color(white)(a/a)|)))

The answer is rounded to two sig figs, the number of sig figs you have for the mass of ethanol.