Test #1 Q. 5

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Jonathan Marcial Dis 1K
Posts: 36
Joined: Fri Apr 06, 2018 11:03 am

Test #1 Q. 5

Postby Jonathan Marcial Dis 1K » Sun May 06, 2018 11:55 am

Maltotriose C18H32O16, 504.437 g/mol is used in the body according to the following overall reaction:

C18H32O16 (s) + 18 O2 (g) --> 18 CO2 (g) + 16 H2o (g)

If you eat something containing 10.0 g of maltotriose, what mass of oxygen must you inhale for the maltotriose to react completely?

What I did was I used n=m/M

so i got n=10g/504.437 g.mol^-1 = .02 mol of C18H32O16

Next I used the same method to try and determine the amount of oxygen needed.

I did .02 mol = m/32 g.mol^-1 and got .64g.

would .64 g of oxygen be the answer for determining how much oxygen one must inhale? If not could someone please help. Thank you!

tmehrazar
Posts: 39
Joined: Fri Apr 06, 2018 11:04 am

Re: Test #1 Q. 5

Postby tmehrazar » Sun May 06, 2018 12:11 pm

I had a different version, but I'm pretty sure the way to do this problem is:

10.0g malrotriose x 1 mol maltotriose/504.437g maltotriose x 18mol O2/1 mol maltotriose x 32g O2/1 mol O2 = 11.42g O2

each denominator for the fractions cancels out the numerator of the previous fraction, so you're left with grams of O2

hope this helps!

Jonathan Marcial Dis 1K
Posts: 36
Joined: Fri Apr 06, 2018 11:03 am

Re: Test #1 Q. 5

Postby Jonathan Marcial Dis 1K » Sun May 06, 2018 12:35 pm

Okay this makes a little more sense than what I was doing. Thank you


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