Posted by Andre on Tuesday, October 19, 2010 at 7:39pm.
Thank you for your help. I did want to clarify why you are only calculating the Molar Mass for H2 and not the molar mass for H2O (or even H2O2). I'm not sure why you did this. Please clarify. See below:
The following reaction represents the decomposition of hydrogen peroxide
2H2O2 -> o2 (g) + 2H2O (l)
How many molecules of water are produced from the decomposition of 3.4g of Hydrogen peroxide, H2O2?
Please walk me through this one, I need more then just the answer, I need to understand how the work was done. I'm very lost at the moment with this question. Thanks in advance for your help.
Chemistry - DrBob222, Tuesday, October 19, 2010 at 5:34pm
Two important things about chemistry.
1. ALWAYS write an equation and balance it.
2. Chemistry works with moles. Therefore, change g to moles first thing you do.
You have the equation. Now change the 3.4 g H2 to moles. moles = grams/molar mass
3.4 g/2.016 = 1.686 moles.
Now using the coefficients in the balanced equation, convert moles H2O2 to moles H2O.
1.686 moles H2 x (2 moles H2O/2 moles H2O2) = 1.686 x (2/2) = 1.686 x (1/1) = 1.686 moles H2O.
Now you know that 1 mole of water is composed of 6.022 x 10^23 molecules. So
1.686 moles H2O x (6.022 x 10^23 molecules H2O/1 mole H2O) = ?? molecules.
Here is a sample stoichiometry problem I've posted. Feel free to use it as needed.
- Chemistry - DrBob222, Tuesday, October 19, 2010 at 8:49pm
Well, it's a long story. First, I misread the problem and worked it initially (and typed all of it out) and discovered before I punched the post answer button that the problem did NOT ask for H2 but for H2O. So I went back and corrected here and there and posted the response; however, I missed some of the corrections I should have made.
mole H2O2 = 3.4/molar mass H2O2 = 3.4/about 34 = 0.1 moles H2O2.
Now convert moles H2O2 to moles H2O.
0.1 mol H2O2 x (2 moles H2O/2 moles H2O2) = 0.1 x (2/2) = 0.1 mol H2O.
Then 0.1 mol H2O x (6.022 x 10^23 molecules H2O/1 mol H2O) = ??
- Chemistry - Andre, Tuesday, October 19, 2010 at 9:09pm
many many many thanks. I can't tell you how long I've been working on figuring this one all out. Your help is very much appreciated not only because you helped answer it but because you helped me understand it.
- Chemistry - DrBob222, Tuesday, October 19, 2010 at 10:16pm
You're very welcome. But you were enough on the ball that you noticed I had goofed and brought it to my attention.
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