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### Quiz 1 Preparation, Question #11

Posted: Sat Jan 28, 2017 11:58 pm
On question #11 of the quiz 1 preparation (pg. 116 in course reader), it asks:
"How many grams of water can be heated from 25.0 degrees Celsius to 100.0 degrees Celsius by the heat released from converting 49.7 g of PbO to Pb?"
The converting reaction is PbO (s) + C (s) --> Pb (s) + CO (g)
delta H= -106.9 kJ

I understand that you would use the equation q= (mass)x(specific heat capacity)x(delta T) and that you can figure out the delta T to be 75.0 degrees Celsius because you are given the two temperatures in the equation. I also know that you find the specific heat capacity of water to be 4.184 because that is given on the formula sheet. However, what I do not understand is that after you calculate the moles of PbO from the 49.7 g of PbO given in the equation, how do you apply that to the q= (mass)x(specific heat capacity)x(delta T) equation in order to find the mass?

### Re: Quiz 1 Preparation, Question #11

Posted: Sun Jan 29, 2017 1:03 am
You can use the $\Delta H$ given to you as a conversion factor of sorts since it is in kJ/mol. So after converting the grams of PbO to moles multiply that value by 106.9 KJ/mol so the moles cancel and you are left with kJ's. This value is the same as q in the equation $q=mC\Delta T$ so you can plug in all the values you know and solve for m, making the mass of water that can be heated 75.8 g. Hope this was helpful!