Posted by Patrick on Wednesday, September 19, 2007 at 4:47pm.
I didnt 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
gg
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
i
i
The measurements are not at STP, they are at 220 celsius and 735 mm Hg. so you use PV=nRT in this instance
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