Question #ffaab

1 Answer
Oct 5, 2015

"0.787 g"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"CH3OH, 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])2CH3OH(l]+3O2(g]2CO2(g]+4H2O(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.