Soccerboi's questions thread

[Tonychet2]

methylpropan-2-ol is better then 2-methylpropan-2-ol since the only place the methyl group can go is on the 2nd carbon so the 2 isnt needed in front of the propanol

same for methylbutane, but i doubt vcaa would take a mark off for writing 2methylbutane or 2methylpropan2ol

edit: sorry didnt read ur question, no probably not

[soccerboi]

1. A mixture of hexane and oxygen was placed in an evacuated vessel at 300°C. The pressure
inside the vessel was 80 kPa. After ignition by a spark, the mixture reacted completely to
form CO2, CO and water vapour. The contents of the container were returned to 300°C and
the pressure was found to be 130 kPa.

Which of the following equations best describes the reaction that took place?
A. C6H14(g) + 8O2(g) → 3CO(g) + 3CO2(g) + 7H2O(g).
B. C6H14(g) + 7O2(g) → 5CO(g) + CO2(g) + 7H2O(g).
C. 2C6H14(g) + 15O2(g) → 8CO(g) + 4CO2(g) + 14H2O(g).
D. 2C6H14(g) + 17O2(g) → 4CO(g) + 8CO2(g) + 14H2O(g).


2. The complete combustion of 24.0 g of an organic compound produces 1.20 mol CO2 and 1.60
mol H2O. The compound was
A. propane.
B. 1-propanol.
C. propanoic acid.
D. methyl ethanoate.

Help please!

[sin0001]

ratio of number of mols of the reactants to the mols of products produced is 8:13 as more pressure means more mols present. So count the number of mols of the reactants and the products of the choices that you have, the ratio should be 8:13. And the correct answer would be B, i think

[sin0001]

Okay so for the second question:
n(C)=n(CO2)=1.2 mol
m(C)= 1.2 x 12 = 14.4 grams of Carbon
n(H)= 2 x n(H2O)= 3.2 mols
m(H)= 3.2 grams of Hydrogen
m(O)= 24 g - (14.4 + 3.2)g= 6.4 grams of oxygen
n(O)= 6.4/16 = 0.4 mols
n(C):n(H):n(O) = 1.2 : 3.2 : 0.4(Divide everything by 0.4) = 3:8:1 (Empirical formula is C3H8O)
Answer is B

[soccerboi]

ratio of number of mols of the reactants to the mols of products produced is 8:13 as more pressure means more mols present. So count the number of mols of the reactants and the products of the choices that you have, the ratio should be 8:13. And the correct answer would be B, i think

Thanks a lot, you got both right!

so is ratios of pressure always equivalent to ration of mols?

[sin0001]

assuming every other variable in pv=nrt stays the same, then i think they should be!

[soccerboi]

This is annoying, i have no idea where to start...

Q) In a 1.00 L vessel, 600 mL of chlorine gas, Cl2, is added to 400 mL of a saturated hydrocarbon that contains two carbon atoms. The vessel is then subjected to UV light for a period of time.

Assuming constant temperature and pressure, what volume is produced of the largest product?
Solution
n(C2H6) : n(Cl2)
1 : 1
So C2H6 is the limiting reagent.
n(C2H6) : n(C2H5Cl)
1 : 1
As all of the species in the equation are all gases, then
V(C2H6) : V(C2H5Cl)
1 : 1
So V(C2H5Cl) = 400 mL

I have no idea what the solutions are on about, someone help me please?

[Phy124]

This is annoying, i have no idea where to start...

Q) In a 1.00 L vessel, 600 mL of chlorine gas, Cl2, is added to 400 mL of a saturated hydrocarbon that contains two carbon atoms. The vessel is then subjected to UV light for a period of time.

Assuming constant temperature and pressure, what volume is produced of the largest product?
Solution
n(C2H6) : n(Cl2)
1 : 1
So C2H6 is the limiting reagent.
n(C2H6) : n(C2H5Cl)
1 : 1
As all of the species in the equation are all gases, then
V(C2H6) : V(C2H5Cl)
1 : 1
So V(C2H5Cl) = 400 mL

I have no idea what the solutions are on about, someone help me please?

Saturated hydrocarbons are alkanes. It has 2 Carbons therefore it must be ethane. The equation between ethane and chlorine gas under those conditions is;

CH₃CH₃ + Cl₂ -> CH₃CH₂Cl + HCl

As temperature and pressure are constant then the ratio of volumes will be the same as the ratio of moles

The volume of ethane is less than that of the chlorine gas, therefore the ethane will be the limiting reactant

The molar ratios for the equation are 1:1:1:1

The chloroethane is the largest molecule (irrelevant because both products will have the same volume I believe)

Due to the molar ratio the chloroethane will have the same amount of moles as the ethane and hence the same volume, which was 400mL.

I think that's right anyway, been a while since I've done much chem

[soccerboi]

Thank you! this makes so much sense!

[soccerboi]

If we react dichromate or permanganate with any compound with a hydroxyl group, does it become a carboxylic acid?

e.g If we oxidise a monosaccharide, disaccharide or polysaccharide does it become a carboxylic acid?

[Somye]

No, because you remember how you must have a primary alcohol to be oxidised to a carboxlic acid? carbohydrates don't have enough bonding positions to be oxidised into a carboxlic acid

[destain]

This is annoying, i have no idea where to start...

Q) In a 1.00 L vessel, 600 mL of chlorine gas, Cl2, is added to 400 mL of a saturated hydrocarbon that contains two carbon atoms. The vessel is then subjected to UV light for a period of time.

Assuming constant temperature and pressure, what volume is produced of the largest product?
Solution
n(C2H6) : n(Cl2)
1 : 1
So C2H6 is the limiting reagent.
n(C2H6) : n(C2H5Cl)
1 : 1
As all of the species in the equation are all gases, then
V(C2H6) : V(C2H5Cl)
1 : 1
So V(C2H5Cl) = 400 mL

I have no idea what the solutions are on about, someone help me please?

Saturated hydrocarbons are alkanes. It has 2 Carbons therefore it must be ethane. The equation between ethane and chlorine gas under those conditions is;

CH₃CH₃ + Cl₂ -> CH₃CH₂Cl + HCl

As temperature and pressure are constant then the ratio of volumes will be the same as the ratio of moles

The volume of ethane is less than that of the chlorine gas, therefore the ethane will be the limiting reactant

The molar ratios for the equation are 1:1:1:1

The chloroethane is the largest molecule (irrelevant because both products will have the same volume I believe)

Due to the molar ratio the chloroethane will have the same amount of moles as the ethane and hence the same volume, which was 400mL.

I think that's right anyway, been a while since I've done much chem

After that the question says what will the total volume of gases present once the reaction has finished.
And since the volume was worked out to be 400ml and for the other product it should be 400ml as well...Then shouldnt that add up to 800ml being present?

[soccerboi]

I think its the total volume of reactants which was 600ml and 400ml is must equal the volume of the products. So you had 600ml and 400ml initially, so end up with 1L at the end.

[ICECOLD]

sorry to intrude, but neap 2012, MC15

how come ethanoic anhydride produces water? :S

[destain]

yeah tahts what was in the answers but from our working out it says that 400ml and 400ml of each gas was produced and therfore it should add up to 800ml what about the other 200ml how would that come in from the working out