Multiple choice (1 Viewer)

[Aerlinn]

Two questions which I figured out I couldn't figure out how to do (they dont look hard, though )

A sample of 1.00g of a pesticide is analysed for its arsenci content by precipitation of the arsenic as the sulfide, As2S3. If 0.123g of the sulfide is obtained, the percentage by mass of arsenic in the pesticide is:
A) 3.75
B) 7.50
C) 37.5
D) 75.0

Equal masses of the two gases oxygen (O2) and sulfur dioxide (SO2) are placed in separate vessels. Both vessels have the same volume and are at the same temperature. The pressure exerted by the oxygen is 100kPa. The pressure, in kPa, exerted by the So2 is closest to:
A) 25
B) 50
C) 100
D) 200

:wave:

 

[TesseracT22]

The first one you can do in c ouple of steps. first find out how many moles of the arsenic sulfide is in 0.123g by dividing this by the molar mass. then multiply this by 2 to find the number of moles of arsenic and then multiply this by arsenics molar mass the find its mass in grams. then just divide this by 1 and multiply by 100 to get a %


In this second one you have to find a ratio of molar mass of So2/molar mass of o2
equate this to x/100 then rearrange to find x which will be your pressure

 

[Aerlinn]

The first one, thanks ... I thought that that's what i did, but maybe I didn't do the calculations properly somewhere

The second one... hm, that would give D as the answer, but the answer's B... =S

~~~
Update: This question...
Ammonia reacts with oxygen to form nitrogen dioxide, NO2, and steam.
57.0L of ammonia is reacted with 140.0L of oxygen at 200C and 4* 10^4Pa
a) which gas does not completely react?
b) what volume of this gas is present is in excess?

I'm wondering if the answer's wrong, since it says that for a) O2 is in excess and for b) 40.25L in excess
I worked out that NH3 was in excess, by 0.5L... (I could be doing it wrong, though...)

 

[morganforrest]

Hey,
Second one isn't so bad... moles=mass/molar mass

So O2 has moles= x/32
and S02 has moles= x/32 + 32 = x/64

Now equal moles of gas occupy an equal volume at the same temp and pressure.

So for this to hold true, the SO2 must have half the pressure as it has twice as many moles.

Hope this helps....if it doesn't post again and I'll try and explain a little better

 

[Aerlinn]

So for this to hold true, the SO2 must have half the pressure as it has twice as many moles.

If no. moles of O2 is x/32 and no. moles SO2= x/64, doesn't that say that SO2 has half as many moles? *ponders*
I am slightly puzzled =S

 

[morganforrest]

no because when you cross multiply the oxygen has 32moles for x grams whereas the sulfur dioxide has 64moles for every x grams.....hence twice as many moles.

 

[TesseracT22]

no because when you cross multiply the oxygen has 32moles for x grams whereas the sulfur dioxide has 64moles for every x grams.....hence twice as many moles.

according to this that means S02 exerts double the pressure of O2 if it has twice as many moles, then shouldnt the answer still be 200 and not 50?

 

[morganforrest]

But go back to the original presumption....equal moles occupy equal volume at same temp and pres.

Since volume is fixed, and temp is fixed the only thing to fiddle with is moles and pressure right??!?!?!!

so if you double the moles you have to halve the pressure to counteract it....ie to balance it out

The equation idea breaks down once you get to the fact that there are twice as many moles and you have to go back to the original presumption. Understanding that is the basis of the question. When you've got figuratively high pressure (like 100kpa for example, that of oxygen), and you have one mole of atoms, they'll be all nice and happy with the distances between the atoms. But if you put twice as many atoms in there then they're all squashed. The way to relieve the PRESSURE of them all being squashed is to halve the pressure to 50 which allows them to spread out more, fulfilling our original presumption.

 

[Aerlinn]

The answer says B, i know that much (of course, it could be wrong...)
I tend to 'sort of' agree with TesseracT22... It a gas has more moles, it probably should exert more pressure. Though i'm not sure i understand how it can be deduced that from no. moles of O2 is x/32 and no. moles SO2= x/64 (since cross multiplying, i wouldn't be able to tell which side was O2 and which side was SO2)... since if you divide the same no. mole by a bigger number, the result should be smaller, and thus no. mole of SO2 should be less...
Hang on, I think I just arrived at the answer ^^

Edit: I think I get what you're saying, morganforrest... though you're talking about relieving the pressure as a result of the 'particles squashed together', though isn't that done to relieve the pressure exerted by the gas particles, which is what we're all getting mixed up over? It is the pressure exerted by each gas that's the relevant idea, I think...

 

[morganforrest]

Oh hang on....I didn't mean cross multiplying the two equations together....just getting rid of the fractions of each of the separate equations ie moving the denominator

 

[morganforrest]

Where was this question from btw?

 

[Aerlinn]

A worksheet from our teacher ^^

just getting rid of the fractions of each of the separate equations ie moving the denominator

You mean, ending up with 2x= x (which should probably be some other letter since they're not the same number)? To me that says... twice the mass of oxygen as there is SO2...? >.<

 

[xiao1985]

Q1) as tesseract said

Q2) it is 50... there are same mass, but SO2 has bigger molar mass, which means SO2 will have lesser moles (1/2 comparing to O2).... at the same volume, O2 will exert twice as much pressure as SO2, hence SO2 will have the pressure half as much as O2

Q3) write the balanced equation first?

 

[yoakim]

This is my working out...

m(O2) = m(SO2)

MM(O2) = 32
MM(SO2) = 64 = 2x more than O2

More MM = Less n

Therefore, as:
MM(SO2)>MM(O2)
n(SO2)<n(O2)

n(SO2) = 1/2 of n(O2)
ie: n(O2) = 2x more than n(SO2)

Thus:
O2 = 2x more pressure than SO2 as no. moles (n) equates to pressure.
ie: the pressure must be halved for SO2.