$B e C_{2} O_{4} \cdot 3 H_{2} O \to B e C_{2} O_{4} (s) + 3 H_{2} O (g)$ $BeC_2O_4 * 3H_2O -> BeC_2O_4 (s) + 3H_2O(g)$ . If 3.21 g of $B e C_{2} O_{4} \cdot 3 H_{2} O$ $BeC_2O_4 * 3H_2O$ is heated to $220^{o} C$ $220^oC$ , how do you calculate the mass of $B e C_{2} O_{4} (s)$ $BeC_2O_4(s)$ formed and the volume of the #H_2O(g) released, measured at 220 C and 735 mm Hg?

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$B e C_{2} O_{4} \cdot 3 H_{2} O \to B e C_{2} O_{4} (s) + 3 H_{2} O (g)$ $BeC_2O_4 * 3H_2O -> BeC_2O_4 (s) + 3H_2O(g)$ . If 3.21 g of $B e C_{2} O_{4} \cdot 3 H_{2} O$ $BeC_2O_4 * 3H_2O$ is heated to $220^{o} C$ $220^oC$ , how do you calculate the mass of $B e C_{2} O_{4} (s)$ $BeC_2O_4(s)$ formed and the volume of the #H_2O(g) released, measured at 220 C and 735 mm Hg?

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Answer 1 · 2018-01-15T10:55:04

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

Explanation:

With respect to $beryllium oxalate trihydrate$ $"beryllium oxalate trihydrate"$ we gots a molar quantity of....

$\frac{3.21 \cdot g}{151.08 \cdot g \cdot m o l^{- 1}} \equiv 0.0212 \cdot m o l$ $(3.21*g)/(151.08*g*mol^-1)-=0.0212*mol$ ...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.... $3*L$

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