#1

So I took chemistry in the summer, but I still have to take a star test late this year. I got a review sheet to work on and I'm stuck on this one question. I don't have a chemistry book so I'm totally lost. Any help (answers or explanations) is appreciated. Thanks.

If 28.7 grams of C6H10 reacts with an excess of oxygen, how many liters of Carbon Dioxide will be formed?

If 28.7 grams of C6H10 reacts with an excess of oxygen, how many liters of Carbon Dioxide will be formed?

#2

27.684

EDIT: only if you add the chlorium benytote²

EDIT: only if you add the chlorium benytote²

#3

only if you add the chlorium benytote²

EDIT: Nevermind you got it

EDIT: Nevermind you got it

#4

1. get your moles of c6h10

2. multiply that by 6 to get your carbon moles, as they are your limitting reagant.

3. plug the result (# of moles of co2) into pv=nrt , I am assuming we must consider STP (273K, 101.something kpa (I'm not sure, we might have to use 760 mmHg, depends on which constant you use for k, it,s been a while since I've done lower level chem, just finsihed organic and am now doing biochem)

4. solve for v, which will give you your volume of co2 obtained.

5. Good luck with the rest, if you have any other queries, shoot.

2. multiply that by 6 to get your carbon moles, as they are your limitting reagant.

3. plug the result (# of moles of co2) into pv=nrt , I am assuming we must consider STP (273K, 101.something kpa (I'm not sure, we might have to use 760 mmHg, depends on which constant you use for k, it,s been a while since I've done lower level chem, just finsihed organic and am now doing biochem)

4. solve for v, which will give you your volume of co2 obtained.

5. Good luck with the rest, if you have any other queries, shoot.

#5

Thanks, Pooch.

#6

If 28.7 grams of C6H10 reacts with an excess of oxygen, how many liters of Carbon Dioxide will be formed?

Ok so:

2C6H10 + 11O2 --> 6CO2 + 10H20

Find the molar mass of C6H10(82.1455) and Divide 28.7 by the molar mass. That will give you the amount of moles of C6H10 present(.343938). Since the ratio is 6 mol of CO2 per 2 mol of C6H10, that means that you multiply .343938 by 3 to get the amount of mols of CO2 that can be made(1.031814). Now that you have the mols of CO2, multiply this by 22.4 to get the vomlume of CO2, which is measured in Liters. The answer is 23.1126336. (Check this I did half of this in my head)

EDIT:

Thats wrong man sorry but you multiply the moles of c6h10 by 3 since 28.7 represnts the amount of 2 moles instead of one once you balance the equation. And all you have to do is multiply the amount of moles of a substance by 22.4 to get the volume in liters.

Ok so:

2C6H10 + 11O2 --> 6CO2 + 10H20

Find the molar mass of C6H10(82.1455) and Divide 28.7 by the molar mass. That will give you the amount of moles of C6H10 present(.343938). Since the ratio is 6 mol of CO2 per 2 mol of C6H10, that means that you multiply .343938 by 3 to get the amount of mols of CO2 that can be made(1.031814). Now that you have the mols of CO2, multiply this by 22.4 to get the vomlume of CO2, which is measured in Liters. The answer is 23.1126336. (Check this I did half of this in my head)

EDIT:

1. get your moles of c6h10

2. multiply that by 6 to get your carbon moles, as they are your limitting reagant.

3. plug the result (# of moles of co2) into pv=nrt , I am assuming we must consider STP (273K, 101.something kpa (I'm not sure, we might have to use 760 mmHg, depends on which constant you use for k, it,s been a while since I've done lower level chem, just finsihed organic and am now doing biochem)

4. solve for v, which will give you your volume of co2 obtained.

5. Good luck with the rest, if you have any other queries, shoot.

Thats wrong man sorry but you multiply the moles of c6h10 by 3 since 28.7 represnts the amount of 2 moles instead of one once you balance the equation. And all you have to do is multiply the amount of moles of a substance by 22.4 to get the volume in liters.

*Last edited by AssassinLoki090 at Feb 20, 2007,*

#7

shit, totally assumed that the molar ratio was 6:1 instead of 3:1, jusst took a fasttrack away from balancing out the equation.