3.21 g BeC2O4*3H2O x (1 mol BeC2O4*3H2O/238.14 g BeC2O4*3H2O) = 0.0134 mol BeC2O4*3H2O
0.0134 mol BeC2O4*3H2O x (3 mol H2O/1 mol BeC2O4*3H2O) = 0.0402 mol H2O
PV = nRT
V = (0.0402 mol H2O x 0.08206 L atm/mol K x 393.15 K)/(735 mm Hg)
V = 0.037 L H2O
If 3.21 g of BeC2O4 * 3H2O is heated to 220 C calculate the volume of the H2O(g) released, measured at 220 C and 735 mm Hg?
please help me
Convert g BeC2O4*3H2O to mols BeC2O4*3H2O.
Convert mols BeC2O4*3H2O to mols H2O.
Use PV = nRT to calculate volume of H2O.
Post your work if you get stuck.
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