Andiio's Chem Questions!

[luken93]

i though it would be :

mass of CO2 = 2.20g

mole of CO2 = mass/Mr = 2.2/44 = 0.050 moles
mole of Carbon = mole of CO2 = 0.050 moles

therefore mass of Carbon = 0.05 x 12 = 0.60g

i guess i cud be wrong ??

Using that method, add up all the components and see if it equals the mass of the ester?

Shouldn't luken's method be correct as he has used the limiting reactant? That's how I would've worked it out.

I thought you couldn't assume that the mole ratio is correct without knowing the full equation and empirical formula of the ester (may or may not be correct).

Yeah I found out the empirical formula of the ester, and then inserted it into the chemical equation before balancing it....

(not saying my approach is correct until it is proven, but it seemed right to me haha)

[luken93]

Yeah I made a mistake, dunno where though...

[david10d]

loledit

[luken93]

mol ratio is incorrect. oxygens don't balance.

yeah fixed it up now

[luken93]

Sammi's version is correct, because m(C) can be worked out purely from the Carbon Dioxide as you know what is in the reaction.

Source: 2002 VCAA Exam 1

[Pixon]

Yeh that's a rather silly question. It's supposed to waste time and trick people into finding the EF (which is more typical question).
luken, you were on the right track, but to avoid the excess oxygen problem you want to find the mass of the carbon and hydrogen from the amounts you calculated. Then subtract those two masses from the 1.02g of ester to find the mass of oxygen and thus the amount of oxygen. Then everything else you did was correct.
If you're curious it's C5H10O2.

[Andiio]

Dw, I worked it out ages ago LOL
THANKS ANYWAYS

[luken93]

Yeh that's a rather silly question. It's supposed to waste time and trick people into finding the EF (which is more typical question).
luken, you were on the right track, but to avoid the excess oxygen problem you want to find the mass of the carbon and hydrogen from the amounts you calculated. Then subtract those two masses from the 1.02g of ester to find the mass of oxygen and thus the amount of oxygen. Then everything else you did was correct.
If you're curious it's C5H10O2.

yeah i realised later on that you could d that, what an annoying q though haha

[Andiio]

Could anyone help me with this please?

Not sure if i'm right

Q: Some rocks were thought to consist of insoluble silica (SiO2) and calcium carbonate (CaCo3). The fraction of CaCO3 in an 8.64g sample of the crushed rock was determined by mixing the sample with excess hydrochloric acid. The acid reacts with CaCO3 according to the following equation.
2HCl(aq) + CaCO3(s) -> CaCl2(aq) + H2O (l) + CO2(g)

b) Excess ammonium oxalate solution was added to the filtered solution. The calcium ions present precipitate as CaC2O4 . H2O.

The CaC2O4 . H2O was collected by filtration, washed and dried. It was then heated to convert it to CaO (molar mass 56.1 g/mol) and a mass of 3.87 g was obtained.

Using this mass of CaO, calculate the percentage of CaCO3 in the rock sample.


Thanks!!

[m@tty]

%%(m/m) 3sf

[Andiio]

Thanks m@tty!

Just another one:

A lawn fertiliser contains mainly ammonium sulfate ((NH4)2SO4) and potassium sulfate (K2SO4). A 1.87 g sample of the fertiliser was dissolved in water and the sulfate precipitated as barium sulfate (BaSO4). The dried BaSO4 had a mass of 2.41g.

a) Write a balanced ionic equation for the precipitation reaction.

Really noob question, but i'm not sure >_<

THanks!

EDIT: Is it just Ba ions + sulfate ions = BaSO4? cept the balancing.. hmm

[Water]

Just another one:

A lawn fertiliser contains mainly ammonium sulfate ((NH4)2SO4) and potassium sulfate (K2SO4). A 1.87 g sample of the fertiliser was dissolved in water and the sulfate precipitated as barium sulfate (BaSO4). The dried BaSO4 had a mass of 2.41g.

a) Write a balanced ionic equation for the precipitation reaction.

Really noob question, but i'm not sure >_<

THanks!

EDIT: Is it just Ba ions + sulfate ions = BaSO4? cept the balancing.. hmm








Somehow, I recognize this problem, from Heineman?
Yup

[Andiio]

Just another one:

A lawn fertiliser contains mainly ammonium sulfate ((NH4)2SO4) and potassium sulfate (K2SO4). A 1.87 g sample of the fertiliser was dissolved in water and the sulfate precipitated as barium sulfate (BaSO4). The dried BaSO4 had a mass of 2.41g.

a) Write a balanced ionic equation for the precipitation reaction.

Really noob question, but i'm not sure >_<

THanks!

EDIT: Is it just Ba ions + sulfate ions = BaSO4? cept the balancing.. hmm








Somehow, I recognize this problem, from Heineman?
Yup

Haha yep!

So it's basically just Ba2+ (aq) + SO4 2- (aq) -> BaSO4 (s) ?

Bit paranoid about any numbers in front since the fertiliser has 2 'S' atoms if you catch my drift :|

[Water]

Yeah lol , I got a bit put off as well but it was just requesting ionic equation so...xD meh Good luck  

PS: The NH4 and The K would be spectators, if you just needed that extra clarity

[Andiio]

Just got another few random questions that have been bugging me.

What is the significant of H+ and H3O+ in acid-base/general chem questions? Does it make much of a difference in an actual question?

Also, if not specified whether the substance is an acid or base, and the pH > 7, do you automatically assume that it is a base?
So say for example an unknown substance attributed a pH of 10; does that mean it's concentration of H+, i.e. C[H+] = 10^-10?