# Question #27cd4

##### 1 Answer

Here's what I got.

#### Explanation:

My guess would be that the question included the **thermochemical equation** that describes the formation of carbon dioxide

#"C"_ ((s)) + "O"_ (2(g)) -> "CO"_ (2(g))" " " "DeltaH_f^@ = -"393.51 kJ mol"^(-1)#

Your goal here is to figure out how much carbon, i.e. how many *grams*, must undergo combustion in order for the reaction to give off the amount of heat needed to heat your sample fo water.

The standard enthalpy of formation, *or* absorbed when **one mole** of a substance is formed from its constituent elements in their most stable form.

In this case,

#DeltaH_f^@ = color(darkorange)(-"393.51 kJ")color(white)(.)color(blue)("mol"^(-1))#

tells you that when **given off**, hence the *minus sign*.

Now, notice that the raction consumes **mole** of carbon to produce **mole** of carbon dioxide, so you can say that **mole** of carbon racts to produce carbon dioxide.

You calculated the heat required to raise the temperature of

#DeltaT = 40^@"C" - 10^@"C" = 30^@"C"#

by using the equation

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

and got

#q = 1000 color(red)(cancel(color(black)("g"))) * "4.18 J" color(red)(cancel(color(black)("g"^(-1)))) color(red)(cancel(color(black)(""^@"C"^(-1)))) * 30color(red)(cancel(color(black)(""^@"C")))#

#q = "125,400 J"#

Now, you know the reaction gives off **for every mole** of carbon that undergoes combustion, so use the amount of heat needed to calculate exactly how many moles would be needed here

#"125,400" color(red)(cancel(color(black)("J"))) * (1 color(red)(cancel(color(black)("kJ"))))/(10^3color(red)(cancel(color(black)("J")))) * "1 mole C"/(393.51color(red)(cancel(color(black)("kJ")))) = "318.67 moles C"#

So, in order to give off **moles** of carbon. To convert this to *grams*, use the **molar mass** of carbon

#318.67 color(red)(cancel(color(black)("moles C"))) * "12.011 g"/(1color(red)(cancel(color(black)("mole C")))) = color(darkgreen)(ul(color(black)("3800 g")))#

I'll leave the answer rounded to two **sig figs**, but keep in mind that your values justify only one significant figure for the answer.