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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 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 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!