# Question 598b4

Oct 19, 2015

Here's what I got.

#### Explanation:

I'll show you how to solve parts (a) and (b) so that you can practice the technique and solve part (c) on your own.

$2 {\text{CH"_3"OH"_text((l]) + color(red)(3)"O"_text(2(g]) -> color(blue)(2)"CO"_text(2(g]) + color(purple)(4)"H"_2"O}}_{\textrm{\left(l\right]}}$

Notice the mole ratios that exist between the species that take part in the reaction. Every 2 moles of methanol will need 3 moles of oxygen gas and produce 2 moles of carbon dioxide and 4 moles of water.

These ratios will be true regardless of how many moles of methanol you start with.

So, how many moles of methanol do you get in your $\text{154 g}$ sample? Use methanol's molar mass to figure that out

154color(red)(cancel(color(black)("g"))) * "1 mole methanol"/(32.042color(red)(cancel(color(black)("g")))) = "4.806 moles methanol"

So, using the aforementioned mole ratios, you will need

4.806color(red)(cancel(color(black)("moles CH"_3"OH"))) * (color(red)(3)" moles O"_2)/(2color(red)(cancel(color(black)("moles CH"_3"OH")))) = "7.209 moles O"_2

of oxygen gas to make sure that all the moles of methanol react. To get the number of grams of oxygen gas, use its molar mass

7.209color(red)(cancel(color(black)("moles O"_2))) * "32.0 g"/(1color(red)(cancel(color(black)("mole O"_2)))) = color(green)("231 g O"_2

Likewise, the reaction would produce

4.806color(red)(cancel(color(black)("moles CH"_3"OH"))) * (color(blue)(2)" moles CO"_2)/(2color(red)(cancel(color(black)("moles CH"_3"OH")))) = "4.806 moles CO"_2

Use carbon dioxide's molar mass to get the number of grams produced

4.806color(red)(cancel(color(black)("moles CO"_2))) * "44.01 g"/(1color(red)(cancel(color(black)("mole CO"_2)))) = color(green)("212 g CO"_2#