Question

a)Calculate the mass percent of carbon in the hydrated form of the solid that has the formula BeC2O4 loosly bonded to 3H2O.

b)When heated to 220degrees C, BeC2O4loosly bonded to 3H2O(s) dehydrates completely as represented below:
BeC2O4loosely bonded to 3H2O ==> BeC2O4 + 3H2O
if 3.21g of BeC2O4loosely bonded to 3H2O is heated to 220 degrees C, calculate
the mass of BeC2O4 formed, and,

the volume of the H2O released, measured at 220 degrees C and 735mm Hg.

For a, there are two carbons for each molecule BeC2O4.3H2O

Percent C= 2*12/molmassBeC2O4.3H2O *100

For b.

Determine the percent of BeC2O4 in the hydrated form.

Percent BeC2O4 = molmassBeC2O4/molmassBeC2O4.3H2O *100

Then grams of BeC2O4 formed is the above percent x 1/100 * 3.21


then how do u find:

the volume of the H2O released, measured at 220 degrees C and 735mm Hg.

hgfkgf
(a)

24.02 g C
---------------------- = .20881 x 100%=
115.0322 g BeC2O4-3H2O

Ans: 20.881%

(b)
(i)
(ii) PV=nRt -> V=nRT
---
P

b.)
i.) 115.028g bec2o4.3h2o
--------------------= 1.1857
97.012gbec2o4

1.1857
------ =3.806
3.21
(3.21g / 151g) x 97g = 2.06g BeC2O4
the 151 is the hydrate's mol mass and the 97 is the BeC2O4 mol mass
a.) (24.02 g C2/151.09 g BeC2O4 + 3H2O) x 100 = 15.9% C2
b) ii. PV=nRT
735 mmHg = .967 atm
220 celsius= 493 K
.0638 mol H2O
.967(V)=.0638*.0821*493
V=26.7 L
no

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