Question #ffaab
1 Answer
Explanation:
The idea here is that you need to use the enthalpy change of reaction for th ecombustion of methanol,
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
Notice that you have
So, how much energy would you need to heat
q = m * c * DeltaT" " , where
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.