Question

Question #40f45

Answer 1

`"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.