Calorimetry calculations

How would I answer this question this is all I’ve done so far.
The answer for q2 is -57kjmol-1
And the answer for q3 is -525kjmol-1.
I don’t understand how they’ve gotten this can someone please explain what I’ve done wrong/need to do.

(edited 2 years ago)

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Reply 1

BlueChicken

Original post by Sarwat.

So, for the first question, your answer is 0.057. Note, this is the actual answer of 57 divided by 1000 - so where do you think the extra factor of 1000 is coming from in your calculation?

Reply 2

Sarwat.

Original post by BlueChicken

So, for the first question, your answer is 0.057. Note, this is the actual answer of 57 divided by 1000 - so where do you think the extra factor of 1000 is coming from in your calculation?

Converting it from j to Kj right because enthalpy change is measured in kj

Reply 3

BlueChicken

Original post by Sarwat.

Converting it from j to Kj right because enthalpy change is measured in kj

That is a true statement and does need to be done. However, are you sure your units are correct in q = mcT?

Reply 4

Turning_A_Corner

Make sure you know which is the limiting reagent as well. Write out the equation and do a molar calculation before you do anything else.

Reply 5

Sarwat.

Original post by BlueChicken

That is a true statement and does need to be done. However, are you sure your units are correct in q = mcT?

yes? I think so..?

Reply 6

Sarwat.

Original post by Turning_A_Corner

Make sure you know which is the limiting reagent as well. Write out the equation and do a molar calculation before you do anything else.

ohhh ok I get it now so you divide the enthalpy change by the limiting reagent?

Reply 7

Turning_A_Corner

Original post by Sarwat.

ohhh ok I get it now so you divide the enthalpy change by the limiting reagent?

The maximum number of moles that can react is determined by the limiting reagent. So the joules/kilojoules calculated by m=c(delta)T will be divided by the moles of the limiting reagent, not the excess reagent.

Reply 8

BlueChicken

Original post by Sarwat.

yes? I think so..?

Look at the units - your mass should be in grams, as the specific heat capacity is given as 4.18 J g-1 K-1 (based on what you have written, you have mass in kg)