# Question ffaab

Oct 5, 2015

$\text{0.787 g}$

#### Explanation:

The idea here is that you need to use the enthalpy change of reaction for th ecombustion of methanol, $\text{CH"_3"OH}$, 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

$\textcolor{red}{2} {\text{CH"_3"OH"_text((l]) + 3"O"_text(2(g]) -> 2"CO"_text(2(g]) + 4"H"_2"O}}_{\textrm{\left(l\right]}}$

You know that the enthalpy change for this reaction is $\Delta H = - \text{1275.8 kJ}$.

Notice that you have $\textcolor{red}{2}$ moles of methanol reacting to give off this much heat. Keep that in mind.

So, how much energy would you need to heat $\text{250.0 g}$ of water from ${20}^{\circ} \text{C}$ to ${35}^{\circ} \text{C}$? Use the equation

$q = m \cdot c \cdot \Delta T \text{ }$, where

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

Plug in your values to get

$q = 250.0 \textcolor{red}{\cancel{\textcolor{b l a c k}{\text{g"))) * 4.18"J"/(color(red)(cancel(color(black)("g"))) * color(red)(cancel(color(black)(""^@"C")))) * (35.0 - 20.0)color(red)(cancel(color(black)(""^@"C}}}}$

$q = \text{15,675 J}$

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.