Question

Question #ffaab

Answer 1

`"0.787 g"`

Explanation:

The idea here is that you need to use the enthalpy change of reaction for th ecombustion of methanol, `"CH"_3"OH"`, to figure out how much methanol would produce enough energy to heat that much water.

The chemical equation for the combustion of methanol looks like this

`color(red)(2)"CH"_3"OH"_text((l]) + 3"O"_text(2(g]) -> 2"CO"_text(2(g]) + 4"H"_2"O"_text((l])`

You know that the enthalpy change for this reaction is `DeltaH = -"1275.8 kJ"`.

Notice that you have `color(red)(2)` moles of methanol reacting to give off this much heat. Keep that in mind.

So, how much energy would you need to heat `"250.0 g"` of water from `20^@"C"` to `35^@"C"`? Use the equation

`q = m * c * DeltaT" "`, where

`m` - the mass of water;
`c` - the specific heat of water, equal to `4.18 "J"/("g" ""^@"C")`;
`DeltaT` - the change in temperature.

Plug in your values to get

`q = 250.0color(red)(cancel(color(black)("g"))) * 4.18"J"/(color(red)(cancel(color(black)("g"))) * color(red)(cancel(color(black)(""^@"C")))) * (35.0 - 20.0)color(red)(cancel(color(black)(""^@"C")))`

`q = "15,675 J"`

Convert this to kilojoules to get

`q = 15,675color(red)(cancel(color(black)("J"))) * "1 kJ"/(10^3color(red)(cancel(color(black)("J")))) = "15.675 kJ"`

So, how many moles would produce this much heat?

`15.675color(red)(cancel(color(black)("kJ"))) * ("2 moles CH"""_3"OH")/(1275.8color(red)(cancel(color(black)("kJ")))) = "0.02457 moles CH"""_3"OH"`

Use methanol's molar mass to determine how many grams would contain this many moles

`0.02457color(red)(cancel(color(black)("moles"))) * "32.04 g"/(1color(red)(cancel(color(black)("mole")))) = color(green)("0.787 g")`

The answer is rounded to three sig figs.