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### L7 Help

Posted: Wed Oct 04, 2017 4:42 pm
For Fundamental L problem 7 why do you have to use 454g of fat for part b. I understand why it is used in part A but i dont think it makes sense to use it while doing stoichiometry for part b.

### Re: L7 Help

Posted: Wed Oct 04, 2017 5:10 pm
Hi! So you're using the 454 grams of fat in part (b) because the question asks "What mass of oxygen is needed to oxidize this amount of tristearin?" and the fat mentioned in the solutions manual is the fat introduced in the intro of the question as tristearin. So the question is actually asking you how much oxygen is needed to oxidize the fat, which is why you start with the mass of the fat given, find the moles of fat, and proceed from there.

Hope that helps!

### Re: L7 Help

Posted: Wed Oct 04, 2017 5:22 pm
In my opinion, L7 (b) states "What mass of oxygen is needed to oxidize this amount of tristearin?" Since they're asking for the same amount (454g), you should still continue using 454g as the amount of tristearin to solve the problem.

### Re: L7 Help

Posted: Wed Oct 04, 2017 10:29 pm
You must use 454g for part (b), because in order to find the mass of oxygen necessary to oxidize the fat, you must relate the moles of C57H110O6 to moles of O2 using their molar ratios from the balanced equation. In order to compare the moles of fat and the moles of oxygen, you must first use the mass of fat (454 g) and divide by its molar mass (891.48 g/mol) to calculate the moles of fat, and then proceed from there. So, in summary:

1) Convert the mass of fat to moles of fat using the molar mass of C57H110O6
2) Multiply the moles of fat by the molar ratio of moles of oxygen : moles of fat
3) Convert the moles of oxygen to mass of oxygen using the molar mass of O2

### Re: L7 Help

Posted: Thu Oct 10, 2019 5:24 pm
What does oxidize mean in this equation? Does it mean simply how much oxygen is needed to react with 454 grams of tristearin?