# Question #f631a

##### 1 Answer

#### Explanation:

The first thing you need to do here is to calculate the amount of heat needed to get **liquid water** at

To do that, use the equation

#color(blue)(ul(color(black)(q = m * c * DeltaT)))#

Here

#q# is the heat gained by the water#m# is themassof the water#c# is thespecific heatof water, equal to#"4.18 J g"^(-1)""^@"C"^(-1)# #DeltaT# is thechange in temperature, defined as the difference between thefinal temperatureand theinitial temperatureof the sample

In your case, you have a mass of

#2 color(red)(cancel(color(black)("kg"))) * (10^3color(white)(.)"g")/(1color(red)(cancel(color(black)("kg")))) = 2 * 10^3# #"g"#

and a change in temperature of

#DeltaT = 100^@"C" - 25^@"C" = 75^@"C"#

which means that you'll have

#q = 2 * 10^3 color(red)(cancel(color(black)("g"))) * "4.18 J" color(red)(cancel(color(black)("g"^(-1)))) color(red)(cancel(color(black)(""^@"C"^(-1)))) * 75color(red)(cancel(color(black)(""^@"C")))#

#q = 6.27 * 10^2 * overbrace(10^3 color(white)(.)"J")^(color(blue)("= 1 kJ"))#

#q = 6.27 * 10^2# #"kJ"#

Now, you know that the enthalpy change of combustion for methane is equal to

#DeltaH_"comb" = - "890 kJ mol"^(-1)#

This means that when **mole** of methane undergoes combustion, **given off**, hence the *minus sign* used for the value of

You can thus say that in order to give off

#6.27 * 10^2 color(red)(cancel(color(black)("kJ"))) * "1 mole CH"_4/(890color(red)(cancel(color(black)("kJ")))) = "0.7045 moles CH"_4#

To convert this to *grams*, use the **molar mass** of methane

#0.7045 color(red)(cancel(color(black)("moles CH"_4))) * "16.05 g"/(1color(red)(cancel(color(black)("mole CH"_4)))) = color(darkgreen)(ul(color(black)("11 g")))#

I'll leave the answer rounded to two **sig figs**, but keep in mind that you only have one significant figure for the mass of water and for the final temperature.