#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
Jan 15, 2018

Answer:

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#