drogen peroxide, H2O2, is unstable under certain conditions and decomposes as follows:

Question

drogen peroxide, H2O2, is unstable under certain conditions and decomposes as follows:

2 H2O2 → 2 H2O + O2

? g 34.00 g 15.98 g

What mass of hydrogen peroxide, to the nearest hundredth of a gram, is decomposed in this reaction?

Answer 1

If you got 34 grams of H2O and 15.98 grams of O2

then add 34 + 15.98
Mass is conserved in the reaction.You have the same total mass on the left as on the right.

Answer 2

Yes, 49.98 SHOULD BE THE CORRECT ANSWER; however, if you ask the problem the other way, that is, starting with 49.98 g H2O2 how much H2O and O2 will you get, here are those numbers.

Start with 49.98 g H2O2 and you will get 26.46 g H2O and 23.52 g O2. That adds up to 49.98 g, as it should, but it isn't 34.00 and 15.98 in the problem. So the problem is flawed or some typos have happened somewhere. It COULD BE that the author of the problem made an error when writing the question. That happens sometimes.

Answer 3

To determine the mass of hydrogen peroxide decomposed in the reaction, we need to use the stoichiometry of the reaction.

The balanced equation tells us that for every 2 moles of hydrogen peroxide (H2O2), we get 2 moles of water (H2O) and 1 mole of oxygen gas (O2).

First, we need to calculate the molar mass of hydrogen peroxide (H2O2):
- The molar mass of hydrogen (H) is 1.008 g/mol.
- The molar mass of oxygen (O) is 16.00 g/mol.
- Since there are two hydrogen atoms and two oxygen atoms in hydrogen peroxide, the molar mass would be: (2 * 1.008 g/mol) + (2 * 16.00 g/mol) = 34.03 g/mol (rounded to two decimal places).

Now let's use the molar mass to calculate the moles of hydrogen peroxide in the given mass of water (34.00 g):
- First, divide the mass of water (34.00 g) by its molar mass (18.02 g/mol) to get the moles of water.
- So, moles of water = 34.00 g / 18.02 g/mol ≈ 1.884 mol (rounded to three decimal places).
- Since the reaction is 2 moles of H2O2 to 2 moles of H2O, the moles of hydrogen peroxide decomposed would be half of the moles of water.
- Therefore, moles of hydrogen peroxide = 1.884 mol / 2 = 0.942 mol (rounded to three decimal places).

Finally, we can calculate the mass of hydrogen peroxide decomposed by multiplying the moles of hydrogen peroxide by its molar mass:
- Mass of hydrogen peroxide = 0.942 mol * 34.03 g/mol ≈ 32.03 g (rounded to two decimal places).

Therefore, the mass of hydrogen peroxide decomposed in this reaction is approximately 32.03 grams.

Answer 4

You need to do a better job of asking your question. It's unclear to me exactly what you have but I THINK you meant for the question mark to be grams H2O2, 34.00 g to be the amount of H2O formed and 15.98 to be the grams of O2 formed. Assuming that is so, remembering the Law of Conservation of Mass, it says that you can't get more out of a reaction than you started with. So you must have had 34.00 g + 15.98 g O2 = 49.98 grams grams H2O2 initially; however, I can tell you that is not the correct answer BECAUSE starting with 49.98 g H2O2 you DO NOT end up with those quantities of H2O and O2. The problem has a flaw in it or you typed it wrong.