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.
In the given question, you correctly used the ideal gas law equation, n = PV/RT, to find the number of moles of gas produced during the decomposition of water. However, it seems you made a mistake in determining the ratio of oxygen to hydrogen.
The balanced chemical equation for the decomposition of water is:
2H2O(l) → 2H2(g) + O2(g)
From this equation, we can see that for every 2 moles of water decomposed, we get 1 mole of O2 and 2 moles of H2. Therefore, the correct ratio of O2 to H2 should be 1:2, not 1:1.
Given that you have calculated 0.327 mol of O2 and 0.654 mol of H2, you have actually found the moles of oxygen correctly, but the moles of hydrogen should be half that value. So, you have 0.654/2 = 0.327 moles of hydrogen.
To find the mass of water decomposed, you need to consider the molar mass of water, which is approximately 18 g/mol. Since the ratio of moles of water to moles of hydrogen is 1:2, the mass of water decomposed would be twice the moles of hydrogen. Therefore, the correct mass of water decomposed would be 2 * 0.327 * 18 = 11.772 grams.
Therefore, the correct answer is indeed approximately 11.8 grams of water that must decompose to yield 24.0 L of gases at 1.00 atm and 25 degrees C.
To determine the number of grams of water that must decompose to yield a given amount of gases, you correctly used the ideal gas law equation n = PV/RT to find the number of moles of gases produced.
Since you obtained a result of 0.981 moles of gases, you then proceeded to find the ratio of oxygen (O2) to hydrogen (H2) based on the stoichiometry of the reaction. According to the balanced chemical equation for the decomposition of water:
2H2O → 2H2 + O2
You correctly deduced that for every mole of oxygen gas produced, there are 0.327 moles of oxygen and 0.654 moles of hydrogen gas produced.
However, it seems that there may be a misunderstanding regarding how to calculate the mass of water decomposed. The correct approach is to use stoichiometry to relate the number of moles of water to the number of moles of oxygen gas produced.
From the balanced chemical equation, we can see that for every mole of water decomposed, we obtain 1 mole of oxygen gas. Therefore, the number of moles of water decomposed is also 0.327 moles.
To determine the mass of water decomposed, we need to use the molar mass of water. The molar mass of water (H2O) is approximately 18.015 g/mol. Therefore, the mass of water decomposed is:
mass = moles * molar mass
mass = 0.327 mol * 18.015 g/mol
mass ≈ 5.89 g
So, the correct answer should be approximately 5.89 g of water that must decompose to yield the given amount of gases. It seems that the book's answer of 11.8 g for each gas is incorrect.
It's important to note that the stoichiometry of a reaction provides the molar ratio between reactants and products, but it doesn't necessarily mean you add the masses of the individual products together. The mass of each product depends on their respective molar masses and the stoichiometric coefficients in the balanced equation.