Question #2cc3c

1 Answer
Jul 7, 2017

Here's what I got.

Explanation:

The idea here is that the heat given off by the combustion of the sample will be equal to the heat absorbed by the calorimeter.

The calorimeter has a heat capacity of "3024 J" ""^@"C"^(-1)3024 JC1, which means that it takes "3024 J"3024 J of heat to increase its temperature by 1^@"C"1C.

In your case, the temperature of the calorimeter increased by 3.337^@"C"3.337C, which implies that the calorimeter absorbed

3.337 color(red)(cancel(color(black)(""^@"C"))) * "3024 J"/(1color(red)(cancel(color(black)(""^@"C")))) = "10,091.1 J"

Convert this to kilojoules

"10,091.1" color(red)(cancel(color(black)("J"))) * "1 kJ"/(10^3color(red)(cancel(color(black)("J")))) = "10.091 kJ"

Now, you know that the calorimeter absorbed "10.091 kJ" of heat, so it must mean that the reaction gave off "10.091 kJ" of heat.

Moreover, you know that this much heat was given off when "0.4075 g" of magnesium underwent combustion. You can thus say that the heat released when "1 g" of magnesium undergoes combustion is equal to

1 color(red)(cancel(color(black)("g"))) * "10.091 kJ"/(0.4075 color(red)(cancel(color(black)("g")))) = "24.76 kJ"

Since this heat is being given off, you can say that the enthalpy change of combustion of magnesium will be equal to

DeltaH_"comb" = color(darkgreen)(ul(color(black)(- "24.76 kJ g"^(-1))))

The minus sign is used to symbolize heat given off.

Finally, to convert this to kilojoules per mole, use the molar mass of magnesium

-24.76 color(white)(.)"kJ"/color(red)(cancel(color(Black)("g"))) * (24.305 color(red)(cancel(color(Black)("g"))))/("1 mole Mg") = color(darkgreen)(ul(color(black)(-"601.8 kJ mol"^(-1))))

The answers are rounded to four sig figs.