Practice Problem

Michelle Steinberg2J
Posts: 73
Joined: Fri Sep 29, 2017 7:04 am

Practice Problem

I came across this practice problem and I'm not sure how to do it. Some help would be great!

How many grams of water can be heated from 25.0 celsius to 100 celsius by the heat released from converting 49.7 g of PbO to Pb?
The converting reaction is: PbO(s) + C(s) yields Pb(s) + CO(g) delta H = -106.9 kJ

Renee Delamater 2H
Posts: 31
Joined: Wed Nov 16, 2016 3:02 am

Re: Practice Problem

I would also like to know how to do this problem! Thanks!

Jingyi Li 2C
Posts: 56
Joined: Fri Sep 29, 2017 7:06 am

Re: Practice Problem

From the reaction, delta H = -106.9 kJ is per mole of PbO. You need to first find the number of moles of 49.7g PbO and then multiply that number by deltaH to get the total heat supplied to the water. Then you divide that amount of heat by the specific heat capacity of water (4.18J/g*°C) and deltaT (100 celsius - 25 celsius) to get how many grams of water can be heated. The basic equation is q = Csp*m*deltaT

Angela 1K
Posts: 80
Joined: Fri Sep 29, 2017 7:05 am

Re: Practice Problem

I'm not sure where you got this problem from, but I just did it through and wanted to check my answers to see if it was correct. I got 75.84 g H2O.

My work is as follows:
$49.7g PbO \cdot \frac{1 mol PbO}{223.2 g PbO}\cdot \frac{106.9 kJ}{1 mol PbO}=23.8 kJ$
$q=mc\Delta T$
$-23,800J=m(4.184 J/gC)(75C)$
$m=75.84 g H2O$