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.
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?
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