Question

<b>(a)calculate the mass percent of carbon in the hydrated form of the solid that has the formula BeC2O4 X 3 H2O (s).</b>

I divided 24.02 g by 151.08 g and got .16, or 16%. is that correct?

Also, I'm having trouble with this follow up question.

<b>When heated to 220 degrees C, BeC2O4 X 3 H2O dehydrates completely as respresented: BeC2O4 X 3 H2O (s)--> BeC2O4 + 3H20 (g). If 3.21 g of BeC2O4 X 3 H2O is heated to 220 degrees C, calculate the:

(i) mass of BeC2O4 (s) formed
(ii) volume of H2O (g) released, measured at 220 degrees C and 735 mm Hg</b>

Answer 1

I didn't check the formula weight of the BeC2O4.3H2O for the first, but the procedure is right.

On the second, calculate the percent composition of 3H2O in the BeC2O4.2H2O, then you can do i easily.

for ii, change it to moles, then V= nRT/P

Answer 2

Yes, the answer comes out to be 15.9%. As for the second part, it is a simple problem.

3.21g BeC2O4 3H2O x (1 mol BeC2O4 3H2O/151.03g) x (1 mol BeC2O4/1 mol BeC2O4) x (97.01/1 mol BeC2O4) = 2.06 g

ii. Since it is measured at STP, we can assume that 22.4 L of the substance = 1 mol.

3 mol H2Ox (22.4 L H2O/1 mol H2O) = 67.2 L H20

Answer 3

The measurements are not at STP, they are at 220 celsius and 735 mm Hg. so you use PV=nRT in this instance