Test 1 Question

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SajaZidan_1K
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Joined: Wed Sep 18, 2019 12:15 am

Test 1 Question

Postby SajaZidan_1K » Sat Nov 02, 2019 9:15 pm

Here is the question: " If 10.00 g of glucose (C6H12O6) is initially present and 5.00 g of H20 is formed, how much grams of glucose is left over? How many moles of 02 were initially present? (glucose is 180.156 g/mol)

I converted water and glucose into moles first, but then I got stuck. What next steps should I take? I did ensure the equation was balanced before I began.

Hannah Pham
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Joined: Fri Aug 09, 2019 12:17 am

Re: Test 1 Question

Postby Hannah Pham » Sat Nov 02, 2019 9:50 pm

You would convert the moles of water formed to moles of glucose using the balanced equation, then convert the amount you found to grams of glucose. After, you would subtract that from the grams of glucose initially present. To determine how many moles of O2 initially present, use the moles of H20 produced to convert to moles of O2 using the balanced equation.

Haley Pham 4I
Posts: 51
Joined: Fri Aug 09, 2019 12:16 am

Re: Test 1 Question

Postby Haley Pham 4I » Sat Nov 02, 2019 10:13 pm

There is also another way you could solve the problem. In order to determine how many grams of glucose is left over, you would first need to determine how many grams of water the 10.0g of glucose produces. You would convert the grams of glucose that was initially present into moles. Then, you would convert the moles of glucose into moles of water using the balanced equation. After, you would that amount into grams so that you can subtract 5.00g of water produced from the grams of water you just calculated to calculate how much excess water would have been yielded by the 10.00g of glucose. Convert the amount of excess water from grams to moles, then convert the moles of water into moles of glucose, then into grams of glucose.
To find out how many moles of O2 were initially present, convert the 5.00g of H2O produced into moles and then into moles of O2 using the balanced equation.

DesireBrown1J
Posts: 98
Joined: Wed Sep 18, 2019 12:18 am

Re: Test 1 Question

Postby DesireBrown1J » Tue Nov 05, 2019 6:13 pm

So for the moles of O2 initially present, did everyone get about 0.28 moles of O2?

DesireBrown1J
Posts: 98
Joined: Wed Sep 18, 2019 12:18 am

Re: Test 1 Question

Postby DesireBrown1J » Tue Nov 05, 2019 6:24 pm

Haley Pham 4I wrote:There is also another way you could solve the problem. In order to determine how many grams of glucose is left over, you would first need to determine how many grams of water the 10.0g of glucose produces. You would convert the grams of glucose that was initially present into moles. Then, you would convert the moles of glucose into moles of water using the balanced equation. After, you would that amount into grams so that you can subtract 5.00g of water produced from the grams of water you just calculated to calculate how much excess water would have been yielded by the 10.00g of glucose. Convert the amount of excess water from grams to moles, then convert the moles of water into moles of glucose, then into grams of glucose.
To find out how many moles of O2 were initially present, convert the 5.00g of H2O produced into moles and then into moles of O2 using the balanced equation.


Did you get 1.67g of glucose left over as your answer?

Justin Ko 4C
Posts: 17
Joined: Mon Sep 23, 2019 12:17 am

Re: Test 1 Question

Postby Justin Ko 4C » Tue Nov 05, 2019 7:46 pm

C6H12O6+6O2-->6H2O+6CO2
10gC6H12O6*(1molC6H12O6/180.156g)*(6molH2O/1molC6H12O6)=0.333molH2O*(18.02gO=H2O/1molH2O)
=6.002gH2O;
6.002gH2O-5.000gH2O=1.002gH2O*(1molH2O/18.02gH2O)
=0.056molH2O excess;
0.056molH2O*(1molC6H12O6/6molH2O)*(180.156gC6H12O6/1molC6H12O6)
=1.675gC6H12O6 leftover

6gH2O*(1molH2O/18.02gH2O)*(6molO2/6molH2O)
=0.278molO2 initially present


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