posted by TJ on .
Water can be decomposed to hydrogen and oxygen. How many grams of water must decompose to yield 24.0L og gases at 1.00atm and 25 degrees C?
I used n=PV/RT to find 0.981 moles of gases. I then took the ratio of O2 to H2 to find the moles of oxygen compared to hydrogen. 0.327 mol to .654 mol. I then used stoichiometry to find the grams of water. 11.8 g each. The book says that is the answer. Why don't I add them?
You added at the beginning so you don't add twice.
2H2O ==> 2H2 + O2
You have 0.327 moles O2 and 0.654 moles H2. Convert each to grams H2O to see what you get.
0.327 x (2 moles H2O/1 mole O2) = 0.327 x 2 = 0.654 moles H2O and that x molar mass = 11.8 g H2O
OR use H2.
0.654 moles H2 x (2 moles H2O/2 moles H2) = 0.654 mole H2O and that x 18 = 11.8.
When you start with 11.8 g H2O, you will produce (and I will use more significant figures than allowed because we rounded the answer from 11.77) 11.77 gH2O x (1 mole H2O/18 g H2O) x (22.4L/1 mol) = 14.647 L H2 at STP and correct that to 25C [14.647 x (298/273)] = 16L H2.
For oxygen we have
11.77 gH2O x (1 mol H2O 18 g H2O) x (1 mole O2/2 mole H2O) x (22.4L/mol) = 7.32L L at STP and corrected for 298 is 7.32 x (298/273) = 8 L O2.
The idea is that by taking 11.8 g H2O you obtain BOTH H2 and O2 at the same time for a combined total of 24.0 L.
I hope I didn't analyze the problem to death.