Question #32eca
1 Answer
Explanation:
The problem tells you that methyl alcohol, which I'll refer to as methanol from this point on,
The first thing to do here is write a balanced chemical equation for this combustion reaction. Start with the unbalanced chemical equation which has methanol and oxygen as reactants and carbon dioxide and water as products
#"CH"_3"OH"_text((l]) + "O"_text(2(g]) -> "CO"_text(2(g]) + "H"_2"O"_text((l])#
In order to have a balanced chemical equation, you must have equal numbers of atoms of each element on both sides of the equation.
Look at carbon. You have one carbon atom on the reactants' side and one on the products' side, so you don't need to make any changes.
Next, focus on hydrogen. You have
#"CH"_3"OH"_text((l]) + "O"_text(2(g]) -> "CO"_text(2(g]) + 2"H"_2"O"_text((l])#
Finally, look at the oxygen atoms. You have
If you multiply the oxygen molecule by
#"CH"_3"OH"_text((l]) + 3/2"O"_text(2(g]) -> "CO"_text(2(g]) + 2"H"_2"O"_text((l])#
To get rid of the fractional coefficient, multiply everything by
#2"CH"_3"OH"_text((l]) + color(red)(3)"O"_text(2(g]) -> 2"CO"_text(2(g]) + color(blue)(4)"H"_2"O"_text((l])#
Now, you know that this reaction consumes
This mole ratio tells you that for every
So, if
#6 color(red)(cancel(color(black)("moles O"_2))) * (color(blue)(4)color(white)(a)"moles H"_2"O")/(color(red)(3)color(red)(cancel(color(black)("moles O"_2)))) = color(green)("8 moles H"_2"O")#