HW Problem: L7 (b)

Moderators: Chem_Mod, Chem_Admin

Ananta3G
Posts: 62
Joined: Wed Sep 18, 2019 12:19 am

HW Problem: L7 (b)

Postby Ananta3G » Mon Sep 30, 2019 2:31 pm

The camel stores the fat tristearin, C57H110O6, in its hump. As well as being a source of energy, the fat is also a source of water because,when it is used, the reaction 2C57H110O6(s)+163O2(g)→114CO2(g)+110H2O(l)2 takes place. (a) What mass of water is available from 1.00 pound (454 g) of this fat? (b) What mass of oxygen is needed to oxidize this amount of tristearin?

Hi - I did part a just fine, but I keep getting the wrong answer for part b. I am using the ratio of 2:163 moles to find out the moles of O2 needed and using the molar mass to get the actual mass in grams but it is not working out. Thank you so much!

Mashkinadze_1D
Posts: 87
Joined: Sat Aug 24, 2019 12:15 am

Re: HW Problem: L7 (b)

Postby Mashkinadze_1D » Mon Sep 30, 2019 2:44 pm

Hello to solve this problem you would proceed as follows:
454g tristearin x 1mol/891.48g = 0.509 mol tristearin

0.509 mol tristearin x 163 mol O2/2mol tristearin x 32gO2/1molO2 = 1327.472 g of O2 but to correct sig figs is 1.33X 10^3g O2.

The important part of this problem is using dimensional analysis to make sure all of the units cancel until you are left with the mass of oxygen. Hope this helps!


Return to “Limiting Reactant Calculations”

Who is online

Users browsing this forum: No registered users and 1 guest