Question #2c18b

For starters, write a balanced chemical equation that describes the combustion of methanol, #"CH"_3"OH"#

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

Notice that every mole of methanol that undergoes combustion produces #color(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 #color(red)(2)# molecules of water.

Your goal now will be to figure out how many molecules of methanol you have in your #"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

#"no. of molecules of H"_2"O" = color(green)(|bar(ul(color(white)(a/a)color(black)(2 * 10^(24))color(white)(a/a)|)))#