#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?
Take the molar quantities....and I gets a volume of under
With respect to
And given the stoichiometry of the reaction....clearly we get
For the volume of water vapour under the given conditions, we solve the old Ideal Gas equation noting that
And I make this approx....