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?

Jan 15, 2018

Take the molar quantities....and I gets a volume of under $3 \cdot L$...

Explanation:

With respect to $\text{beryllium oxalate trihydrate}$ we gots a molar quantity of....

$\frac{3.21 \cdot g}{151.08 \cdot g \cdot m o {l}^{-} 1} \equiv 0.0212 \cdot m o l$...and we dehydrate this material according to the following equation....

$B e {C}_{2} {O}_{4} \cdot 3 {H}_{2} O \left(s\right) \stackrel{\Delta}{\rightarrow} B e {C}_{2} {O}_{4} \left(s\right) + 3 {H}_{2} O \left(g\right) \uparrow$

And given the stoichiometry of the reaction....clearly we get $0.0212 \cdot m o l$ anhydrous beryllium oxalate, and $3 \times 0.0212 \cdot m o l = 0.0637 \cdot m o l$ water vapour, i.e. a mass of $1.15 \cdot g$.

For the volume of water vapour under the given conditions, we solve the old Ideal Gas equation noting that $760 \cdot m m \cdot H g \equiv 1 \cdot a t m$.

$V = \frac{n R T}{P} = \frac{0.0637 \cdot m o l \cdot 0.0821 \cdot \frac{L \cdot a t m}{K \cdot m o l} \cdot 493.15 \cdot K}{\frac{735 \cdot m m \cdot H g}{760 \cdot m m \cdot H g \cdot a t {m}^{-} 1}}$

And I make this approx....$3 \cdot L$