Question

`BeC_2O_4 * 3H_2O -> BeC_2O_4 (s) + 3H_2O(g)`. If 3.21 g of `BeC_2O_4 * 3H_2O` is heated to `220^oC`, how do you calculate the mass of `BeC_2O_4(s)` formed and the volume of the #H_2O(g) released, measured at 220 C and 735 mm Hg?

Answer 1

Take the molar quantities....and I gets a volume of under `3*L`...

Explanation:

With respect to `"beryllium oxalate trihydrate"` we gots a molar quantity of....

`(3.21*g)/(151.08*g*mol^-1)-=0.0212*mol`...and we dehydrate this material according to the following equation....

`BeC_2O_4*3H_2O(s) stackrelDelta rarrBeC_2O_4(s)+3H_2O(g)uarr`

And given the stoichiometry of the reaction....clearly we get `0.0212*mol` anhydrous beryllium oxalate, and `3xx0.0212*mol=0.0637*mol` water vapour, i.e. a mass of `1.15*g`.

For the volume of water vapour under the given conditions, we solve the old Ideal Gas equation noting that `760*mm*Hg-=1*atm`.

`V=(nRT)/P=(0.0637*mol*0.0821*(L*atm)/(K*mol)*493.15*K)/((735*mm*Hg)/(760*mm*Hg*atm^-1))`

And I make this approx....`3*L`