coletrain chemistry question

[jay1993]

I just chose to use S.T.P (standard temp and pressure) where 1 mol = 24.5 L

Is that the part they left out in the question? I mean without assuming STP the Q is like impossoble to know (For me anyway)....

[stonecold]

I just chose to use S.T.P (standard temp and pressure) where 1 mol = 24.5 L

Is that the part they left out in the question? I mean without assuming STP the Q is like impossoble to know (For me anyway)....

Nah, they don't have to put it in.

You could use SLC (standard lab conditions) where 1 mol = 22.4 L and it still should work.

[jay1993]

But they just said pressure and temperature reamined the same... in the exam are they more specific?

ohh no i see as long as u get a mole ratio u can figure out the molecular formula?

[stonecold]

^Yeah, so if you use SLC instead, the mole ratio is still 6.5.

An easier way to think of it, is that when Temp and Pressure are the same, Volume is equal to Moles.

This is true becaue 6.5 L: 1 L is the same as the calculated mole ratio 6.5 mol : 1 mol.

[Mao]

This was a multiple choice question

5.00g of a mixture of magnesium and zinc was allowed to completely react with dilute sulfuric acid according to the equation;
M + H2SO4----> MSO4 + H2     (M is the mixture of zinc and magnesium)

If .250g of hydrogen gas was produced in the reaction, what would the percentage composition of the magnesium be?

the options are
a) 17.2%     
b) 82.8%
c) 37.6%
d) 62.4%

This question is quite mathematically involved, but not impossible.

n(H2) = 0.250 / 2 = 0.125 mol

Since Mg and Zn both react on a 1:1 ratio with H2SO4 to release 1:1 H2 gas, this implies

n(Mg) + n(Zn) = 0.125 mol  ----------[1]

Also, the mass of Mg and Zn add up to 5.00g

n(Mg) * 24.3 + n(Zn) * 65.4 = 5.00  ------------[2]

We see that's a set of simultaneous equation, multiplying [1] by 24.3 and eliminate

n(Zn) * 41.1 = 1.9625
n(Zn) = 0.04775 mol
m(Zn) = 0.04775 * 65.4 = 3.12g
%(Zn) = 3.12/5.00 * 100% = 62.4%

%(Mg) = 100% - 62.4% = 37.6%
:. Option C

[was this from TSFX?]

[Blakhitman]

Damn that's complicated!

[stonecold]

it's actually quite clever.  i never would have got it though lol.
at least now i know how to do it.

what are the odds of something like this appearing on vcaa exams?

[Mao]

it's actually quite clever.  i never would have got it though lol.
at least now i know how to do it.

what are the odds of something like this appearing on vcaa exams?

Never.

[coletrain]

Thanks Mao

[coletrain]

An impure sample of iron(ii) sulfate, weighing 1.545g was treated to produce a precipitate of Fe2O3. If the mass of the dried precipitate was .315g, calculate the percentage of iron in the sample.

Can someone show me the equations i need for this, cause the original was iron(ii) and the precipitate was iron(iii)

[vexx]

An impure sample of iron(ii) sulfate, weighing 1.545g was treated to produce a precipitate of Fe2O3. If the mass of the dried precipitate was .315g, calculate the percentage of iron in the sample.

Can someone show me the equations i need for this, cause the original was iron(ii) and the precipitate was iron(iii)

Well FeS04 + 02- --> Fe2O3 + S042-
(the other ions aside from the ones with Fe aren't too important)
then balance so 2FeS04 --> Fe203

So find the mol of Fe2O3

Then x2 to get the mol of FeS04

Since it's asking for the percent of IRON, you times the FeS04 x1 to get Fe2+
(basically n(FeS04)=n(Fe) )

Then xMolar Mass of Fe to get the amount in grams

Then it's that dvided by 1.545g x100

= % of iron in sample

[coletrain]

thanks vexx

[coletrain]

another one
when 10.0g of white phosphorus is burned in oxygen a .228g of an oxide of phosphorus is produced. The molar mass is 284g/mol, find the empirical formula?

[vexx]

another one
when 10.0g of white phosphorus is burned in oxygen a .228g of an oxide of phosphorus is produced. The molar mass is 284g/mol, find the empirical formula?

err. are you sure it's 10g that is burnt? i think i did this question and it the amount of phosphorous burned was less then the amount of total oxide of phosphorous produced.

[coletrain]

yeah ur right its .100g my bad