Homework questions thread

[Kaille]

Here you go,

OH-(aq) + H+(aq) ---> H2O(l)
n(NaOH)=n(OH-)=cV=(30/1000)*10=0.3mol
n(HCl)=n(H+)=cV=(20/1000)*0.1=0.002mol
:. OH-(aq) is in excess
n(OH-)excess=n(OH-)original-n(OH-)reacted
n(OH-)reacted=n(H+)reacted=0.002mol
:. n(OH-)excess=0.3-0.002=0.298mol
[OH-]resultant=0.298/((20+30)/1000)=5.96M
pOH=-log[OH-]=-log(5.96)=-0.77525
pH=14-pOH=14+0.77525=14.78
One very basic solution!

that's exactly what i did but the answer is 12.3?

[Panicmode]

Here you go,

OH-(aq) + H+(aq) ---> H2O(l)
n(NaOH)=n(OH-)=cV=(30/1000)*10=0.3mol
n(HCl)=n(H+)=cV=(20/1000)*0.1=0.002mol
:. OH-(aq) is in excess
n(OH-)excess=n(OH-)original-n(OH-)reacted
n(OH-)reacted=n(H+)reacted=0.002mol
:. n(OH-)excess=0.3-0.002=0.298mol
[OH-]resultant=0.298/((20+30)/1000)=5.96M
pOH=-log[OH-]=-log(5.96)=-0.77525
pH=14-pOH=14+0.77525=14.78
One very basic solution!

that's exactly what i did but the answer is 12.3?

I did something different but still got the same answer. Maybe the answer is wrong? :S

[attachment deleted by admin]


EDIT:

Attachments re-attached

[Hamdog17]

Yeah the answer maybe wrong. Thinking about it 10M NaOH is very concentrated so adding a dilute acid to that would still leave the pH very high.

[Panicmode]

May I ask why the attachments were removed? Were they too large/too many? Did I break some rule?

Just wanna know for future reference . Sorry to sidetrack the thread a bit...



@Hamdog Yeah 10M NaOH is really concentrated, like industrial strength.

[ninwa]

I think David removed them because of a glitch which meant that you couldn't view page 5 of this thread.

Try uploading them again? Sorry for the inconvenience!

[Bismuth]

Essentially, you weigh the sample of a compound (usually unknown) and heat the sample in excess oxygen. Combustion occurs, and the products will be CO₂(g) and H₂O(g). The products alongside the unreacted O₂(g) are passed through CaCl₂ which is hygroscopic, and therefore acts as a desiccant. CaCl₂ absorbs H₂O and the mass of H₂O can therefore be determined.

The remaining gases are passed through NaOH(aq), and CO₂(g) is 'absorbed' (it produces carbonates and bicarbonates to my knowledge; doesn't really matter for the purpose of this exercise). The fact is, the mass of the CO₂ can be determined due to the increase in mass. The idea is to determine the empirical formula, and they usually toss in some additional information with these sorts of questions using the general gas law so you can work out the molar mass and then the molecular formula.

[mihir94]

Hi can someone help me find what reactant is in excess in this scenario.

100ml of methane and 100ml of oxygen were mixed together and the mixture was sparked. The eqution for the combustion reaction is:

CH4 (g) + 2O2 (g) ---> CO2 (g) + 2H2O (g)

All gases were measured at the same temperature and pressure. Under these conditions, water is in the gas state.

[Shenz0r]

Since water is in the gas state, it's possible that the conditions are in SLC, so you can work out the amount of each gas through molar volume (in SLC, V_m = 24.5)

Work out the number of mol of CH₄ and O₂ using V_m = V/n.
1. n(CH₄) = 0.1/24.5 = 0.004082, n(O₂) = 0.1/24.5 = 0.004082
From the equation, one mol of CH₄ reacts with 2 mol of O₂. Therefore, 0.004082 mol of CH₄ would react with 0.008163 mol of O₂
But the number of mol of O₂ (0.004082 mol) is less than the 0.008163 mol required, therefore oxygen drives the reaction and is the limiting reagent, and methane would be the excess reagent as some of it remains after the reaction.

I'm pretty sure that's how you do it.

[mihir94]

Since water is in the gas state, it's possible that the conditions are in SLC, so you can work out the amount of each gas through molar volume (in SLC, V_m = 24.5)

Work out the number of mol of CH₄ and O₂ using V_m = V/n.
1. n(CH₄) = 0.1/24.5 = 0.004082, n(O₂) = 0.1/24.5 = 0.004082
From the equation, one mol of CH₄ reacts with 2 mol of O₂. Therefore, 0.004082 mol of CH₄ would react with 0.008163 mol of O₂
But the number of mol of O₂ (0.004082 mol) is less than the 0.008163 mol required, therefore oxygen drives the reaction and is the limiting reagent, and methane would be the excess reagent as some of it remains after the reaction.

I'm pretty sure that's how you do it.

Oh okay. Do you always just presume the conditions are SLC if nothing is stated and that water is in the gas state?

[Aurelian]

Since water is in the gas state, it's possible that the conditions are in SLC, so you can work out the amount of each gas through molar volume (in SLC, V_m = 24.5)

...what? If anything the fact that water is present as a gas should indicate conditions are *not* SLC.

The trick to this question is simple - because temperature and pressure are constant, amount ratios are equivalent to volume ratios.

"100ml of methane and 100ml of oxygen were mixed together and the mixture was sparked. The eqution for the combustion reaction is:

CH4 (g) + 2O2 (g) ---> CO2 (g) + 2H2O (g)"

n(CH4):n(O2) = V(CH4):V(O2) = 1:2

If 100ml of each are mixed together, it becomes clear upon inspection that methane is in excess (100ml methane would require 200ml O2).

[mihir94]

Since water is in the gas state, it's possible that the conditions are in SLC, so you can work out the amount of each gas through molar volume (in SLC, V_m = 24.5)

...what? If anything the fact that water is present as a gas should indicate conditions are *not* SLC.

The trick to this question is simple - because temperature and pressure are constant, amount ratios are equivalent to volume ratios.

"100ml of methane and 100ml of oxygen were mixed together and the mixture was sparked. The eqution for the combustion reaction is:

CH4 (g) + 2O2 (g) ---> CO2 (g) + 2H2O (g)"

n(CH4):n(O2) = V(CH4):V(O2) = 1:2

If 100ml of each are mixed together, it becomes clear upon inspection that methane is in excess (100ml methane would require 200ml O2).

That makes sense. Thank you!!!

[Shenz0r]

Since water is in the gas state, it's possible that the conditions are in SLC, so you can work out the amount of each gas through molar volume (in SLC, V_m = 24.5)

...what? If anything the fact that water is present as a gas should indicate conditions are *not* SLC.

Crap, my bad, my brain really isn't working LOL

[soccerboi]

20g of copper was heated in a current of chloride and then weighed, 30.7g of copper metal and copper(II) chloride being obtained. Calculate the volumn of chlorine used in the experiment, measured at 20 degrees celcius and 763mm Hg. Ans is 3.61dm^3

i used this equation:  2Cu+2Cl--->Cu + CuCl2
but got about double the ans
Have i done soemthing wrong?

n(Cu)=20/63.6= 0.314mol
V(Cl)=(0.314x8.31x293)/101.7
  = 7.518 dm^3

[soccerboi]

But it says "30.7g of copper metal and copper(II) chloride being obtained" so how come the products arent Cu (copper metal) + CuCl2 (copper chloride(II))? and also does the 30.7g refer only to copper metal or the CuCl2?
And also, using that eqn, the mol of Cu( 0.314mol) still equals the mol of Cl, so i still dont get the ans.

[Panicmode]

20g of copper was heated in a current of chloride and then weighed, 30.7g of copper metal and copper(II) chloride being obtained. Calculate the volumn of chlorine used in the experiment, measured at 20 degrees celcius and 763mm Hg. Ans is 3.61dm^3

i used this equation:  2Cu+2Cl--->Cu + CuCl2
but got about double the ans
Have i done soemthing wrong?

n(Cu)=20/63.6= 0.314mol
V(Cl)=(0.314x8.31x293)/101.7
  = 7.518 dm^3

See attached scan .