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Topic: When should I solve for qRXN vs. qH2O/qcal? specifics below.  (Read 269 times)

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Offline knightstar33

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When should I solve for qRXN vs. qH2O/qcal? specifics below.
« on: November 11, 2020, 05:28:34 PM »
My teacher gave me 2 problems:
1. Calculate the heat of dissolution of MgCl2: solved for qRXN
2. Calculate the heat of combustion of C4H10O: solved for qcal
Why? Why am I solving for 2 different types of q when they are essentially the same problem?

For the 2nd problem, here is the background info: 7.00 mL of C4H10O (density = 0.714 g/mL) is burnt in a calorimeter (calorimeter heat capacity = 10.34 kJ/°C). The temperature in the calorimeter rises from 25.0 °C to 39.7 °C.
Then, it asks me to calculate q when 1 mole of C4H10O is burned.
Here's the process to solve for q: qreaction = (74g/mol)/5.00 g x (-152 kJ) = - 2.25 x 10 kJ/mol
I get where 5g, and -152 kJ comes from, and 74g/mol is the molar mass of C4H10O.
But why do I multiply these numbers together? Is it some kind of formula? And, if it asks me to calculate for when 1 mole is burned, then why do I have to do 74/5? Can't I just do 74g because that's 1 mole?

Sorry for the word vomit, I'm just so confused.


Offline chenbeier

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Re: When should I solve for qRXN vs. qH2O/qcal? specifics below.
« Reply #1 on: November 12, 2020, 06:21:19 AM »
First you calculate the mass of 7 ml C4H10O by using the specific gravity of 0,714 g/ml
This gives you m= Rho* V  ~ 5g
This 5 g you convert to mole by using molar mass.
n = m/M
This gives n = 5g/74 g/mol = 0,068 mol
Finally you calculate the q = 152 kJ/0.068 mol  = 2,25* 103 kJ/mol
 If you do all together  then you have q  = 152 kJ/(m/M) = 152 kJ*M/m = 152kJ*74 g/ mol/5 g


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