Question #ffaab

1 Answer
Oct 5, 2015

Answer:

#"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.