#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?

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?

Ellie Andrews · Accepted Answer

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

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#

Avery Adams · Answer

undefined In order to compute the mass of BeC2O4(s) formed, first find the molar mass of BeC2O4*3H2O. Next, use stoichiometry to find the moles of BeC2O4(s) formed. Lastly, use the molar mass of BeC2O4 to convert the moles to grams. The volume of H2O(g) released can be computed using the ideal gas law equation, PV = nRT, where P is pressure, V is volume, n is the number of moles, R is the gas constant, and T is temperature. Convert the given pressure to atm and the given temperature to Kelvin. Next, use stoichiometry to find the moles of H2O(g) released. Lastly, use the molar volume of a gas at STP to convert moles to volume.

#BeC_2O_4 * 3H_2O -&gt; 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?

#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?