# Question 2c18b

Jul 20, 2016

$2 \cdot {10}^{24} \text{molecules of H"_2"O}$

#### Explanation:

For starters, write a balanced chemical equation that describes the combustion of methanol, $\text{CH"_3"OH}$

${\text{CH"_ 3"OH"_ ((l)) + 2"O"_ (2(g)) -> "CO"_ (2(g)) + color(red)(2)"H"_ 2"O}}_{\left(l\right)}$

Notice that every mole of methanol that undergoes combustion produces $\textcolor{red}{2}$ moles of water.

Since a mole of a substance is simply a very, very large collection of molecules of said substance, you can say that every molecule of methanol that undergoes combustion produces $\textcolor{red}{2}$ molecules of water.

Your goal now will be to figure out how many molecules of methanol you have in your $\text{50 g}$ sample. Use methanol's molar mass to convert the mass to moles

50 color(red)(cancel(color(black)("g"))) * ("1 mole CH"_3"OH")/(32.04color(red)(cancel(color(black)("g")))) = "1.56 moles CH"_3"OH"

Next, use Avogadro's number to calculate how many molecules of methanol you have in this many moles

1.56 color(red)(cancel(color(black)("moles CH"_3"OH"))) * overbrace((6.022 * 10^(23)"molec. CH"_3"OH")/(1color(red)(cancel(color(black)("mole CH"_3"OH")))))^(color(blue)("Avogadro's number")) = 9.39 * 10^(23)"molec. CH"_3"OH"

Finally, use the aforementioned molecule ratio to find the number of molecules of water produced by the reaction

9.39 * 10^(23)color(red)(cancel(color(black)("molec. CH"_3"OH"))) * (color(red)(2)color(white)(a)"molec. H"_2"O")/(1color(red)(cancel(color(black)("molec. CH"_3"OH")))) = 1.878 * 10^(24)"molec. H"_2"O"#

Since you only have one sig fig for the mass of methane, you can only have one sig fig for the answer

$\text{no. of molecules of H"_2"O} = \textcolor{g r e e n}{| \overline{\underline{\textcolor{w h i t e}{\frac{a}{a}} \textcolor{b l a c k}{2 \cdot {10}^{24}} \textcolor{w h i t e}{\frac{a}{a}} |}}}$