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?

1 Answer
anor277
Jan 15, 2018

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

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