[jaydee54]

YBa2Cu3O(7−x)(s) + ([ 7/2]−x)H2(g) → [ 1/2]Y2O3(s) + 2BaO(s) +3Cu(s) + ([ 7/2]−x)H2O(g)

35.487 ± 0.005 mg of solid YBa2Cu3O(7−x) reacts completely, according to the above equation, in a stream of hydrogen gas at 1000°C, leaving 32.825 ± 0.005 mg of solid residue.

(For the purpose of this exercise treat the values below as if they were exact.)

Molar mass of YBa2Cu3O(7−x) is (666.194 - 15.9994x)
Mass equivalent of (0.5Y2O3 + 2BaO + 3Cu) is 610.196

Calculate the value and (absolute) uncertainty of x.


I found there is going to be 35.487 - 32.825 = 2.662g of H20
 2.662g / 18.0 g/mol = 0.1479 mol of H20.

That is how far I have got to and am clueless as to what to do next.

[Borek]

I found there is going to be 35.487 - 32.825 = 2.662g of H20

mg not g.

You have to express information given in a form of equation combining combining molar mass of YBa₂Cu₃O_(7−x) and masses of water and YBa₂Cu₃O_(7−x). It gets convoluted, but in general it is about a simple conversion between mass, molar mass, number of moles and about simple use of stoichiometric coefficients. Once you have this equation it will contain just one unknown - x.

[jaydee54]

mol of O in reactants = mol of O in products

(0.035487 / (666.194 - 15.9994x))(7-x) = (0.032825/610.196)(1.5 + 2) + (0.002662/18.0)(3.5-x)

And then solving for x = 3.34 but that apparently is the wrong answer. Am I missing anything in the above equation?

[Borek]

Oxygen balance looks like a correct idea, however

(0.002662/18.0)(3.5-x)

is not a number of moles of oxygen in water produced.

But even after correcting it the result I got differs from the one you would got by about 0.04, so could be there is something else wrong - either on my side, or on your side.

[Borek]

OK, don't worry about the difference between our result, looks like the system is very sensitive to the numerical errors.

Edit: or not?

Sigh, I have no time to play with it.