Homework problem L.7b

Moderators: Chem_Mod, Chem_Admin

Posts: 51
Joined: Fri Sep 29, 2017 7:04 am

Homework problem L.7b

Postby JennyCKim1J » Wed Oct 04, 2017 12:18 am

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 2 C57H110O6(s) + 163 O2(g) -> 114 CO2(g) + 110H2O(l) takes place. What mass of oxygen is needed to oxidize this amount of tristerain?

I did:

(454g fat)/(1/891.44g*mol^-1) = .509 mol fat
(.509 mol fat)/(163 mol O2/2 mol fat) = 41.5 mol O2
(42.51 mol O2)/(32.0g*mol^-1) = 1.33g O2

and got 1.33g O2 as the final answer. However, the answer on the solutions manual is 1.33g O2*10^3. I don't understand why there is 10^3 at the end?

Madelyn Gehrich 1E
Posts: 23
Joined: Fri Sep 29, 2017 7:05 am
Been upvoted: 1 time

Re: Homework problem L.7b

Postby Madelyn Gehrich 1E » Wed Oct 04, 2017 1:18 am

You switched the last part of the equation. You have 42.51 mol O2 but you have to multiply that by ((32.00 g O2)/(1 mol O2)). In your equation you did division instead of multiplication for the grams of oxygen gas. If you fix this, it should give you 1.33 x 10^3 g O2!

Return to “Limiting Reactant Calculations”

Who is online

Users browsing this forum: No registered users and 1 guest