Missed this Question on Test1

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derinceltik1K
Posts: 51
Joined: Sat Aug 17, 2019 12:15 am

Missed this Question on Test1

Postby derinceltik1K » Sat Oct 19, 2019 2:47 pm

If 10.00 g of glucose is initially present and 5.00 g of H2O is formed, how much grams of glucose is left over? How many moles of O2 were initially present? (glucose is 180.156 g/mol)

C6H12O6 + 6O2 -> 6CO2 + 6 H2O + ENERGY

I missed this question on Test 1. Can someone explain how to solve this problem? Thanks

rabiasumar2E
Posts: 108
Joined: Thu Jul 11, 2019 12:15 am

Re: Missed this Question on Test1

Postby rabiasumar2E » Sat Oct 19, 2019 6:38 pm

First I made sure that the equation was balanced.
Then I did the following to find out how much of CO2 and H20 was used to make glucose and therefore figure out what the limiting reagent was.

10.0g CO2 (1mol Co2/ 44.011g CO2) (1mol C6H12O6/ 6mol CO2) (180.156g C6H12O6/ 1 mol C6H12O6) = 6.822g


20.0g H20 (1 mol H20/ 18.016g H20) (1mol C6H12O6/ 6 mol H20) (180.156G C6H12O6/ 1 mol C6H12O6) = 33.33g

making CO2 the limiting reagent and 6.822g of glucose produced.

I'm sorry if this was confusing and I hope it helps.

rabiasumar2E
Posts: 108
Joined: Thu Jul 11, 2019 12:15 am

Re: Missed this Question on Test1

Postby rabiasumar2E » Sat Oct 19, 2019 6:38 pm

First I made sure that the equation was balanced.
Then I did the following to find out how much of CO2 and H20 was used to make glucose and therefore figure out what the limiting reagent was.

10.0g CO2 (1mol Co2/ 44.011g CO2) (1mol C6H12O6/ 6mol CO2) (180.156g C6H12O6/ 1 mol C6H12O6) = 6.822g


20.0g H20 (1 mol H20/ 18.016g H20) (1mol C6H12O6/ 6 mol H20) (180.156G C6H12O6/ 1 mol C6H12O6) = 33.33g

making CO2 the limiting reagent and 6.822g of glucose produced.

I'm sorry if this was confusing and I hope it helps.

Andrew F 2L
Posts: 103
Joined: Sat Aug 17, 2019 12:17 am

Re: Missed this Question on Test1

Postby Andrew F 2L » Sat Oct 19, 2019 6:45 pm

Hey!
So I started by converting Carbon Dioxide and Water into moles from the 10 grams of Carbon Dioxide and 20 grams of water that they tell us is given. Because the balanced reaction calls for 6 moles of each, I divided the number of moles we calculated in both carbon dioxide and water by 6 and the smaller number is going to be the limiting reactant. In this case it is Carbon Dioxide and so now we know that the max amount of glucose we can actually produce is 0.03788 moles of Glucose and you just need to convert this to grams by the molar mass of Glucose to get 6.824 grams of Glucose

Jessica Chen 2C
Posts: 103
Joined: Thu Jul 11, 2019 12:17 am

Re: Missed this Question on Test1

Postby Jessica Chen 2C » Sun Oct 20, 2019 10:44 am

So I looked back at my test to check what I did for this problem, and I realized that I can't seem to find this question on my test. I guess this means there's different versions of the test out there? (Also, based on what you've said for the question, I think you could also convert from 5 grams of H20 to grams of glucose using stoichiometry, then subtract the number you get from the original 10.00 grams to see how many grams of glucose you have left over. To find moles of O2, once again convert from 5g H20 to moles of O2 using stoichiometry, then you'll get your answer.)

derinceltik1K
Posts: 51
Joined: Sat Aug 17, 2019 12:15 am

Re: Missed this Question on Test1

Postby derinceltik1K » Mon Oct 21, 2019 11:40 pm

Yes there are different versions of the test for each TA's discussion class I believe. Thanks for the answer though.


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