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Chemistry 10 Online
OpenStudy (tywower):

How many liters of oxygen gas can be produced if 28.7 grams of water decomposes at 294 Kelvin and 0.986 atmospheres? Show all of the work used to solve this problem. 2 H2O (l) yields 2 H2 (g) + O2 (g)

OpenStudy (tywower):

@Abhisar

OpenStudy (tywower):

@sweetburger

OpenStudy (tywower):

@paki

OpenStudy (tywower):

@Kbug

OpenStudy (abhisar):

Have you tried to do it on your own?

OpenStudy (tywower):

no, im having trouble remembering how to do this

OpenStudy (tywower):

@Abhisar

OpenStudy (abhisar):

First you need to find the number of moles of oxygen formed. We can see that 1 mole of water forms 0.5 moles of oxygen so 28.7 grams of water i.e 28.7/18 =1.6 moles of water will form 0.8 moles of water. Any problem yet?

OpenStudy (abhisar):

Now use the universal Gas law PV=nRT to find the volume of oxygen, Pressure is given 0.986 atm, T= 294 K V=nRT/P

OpenStudy (tywower):

no problem yet @Abhisar

OpenStudy (tywower):

1.01 times 2 plus 16.00 equals 18.02 grams per mole 28.7 grams of H2O times (1 mole divided by 18.02 grams) equals 1.59 moles of H2O water:oxygen = 2:1 is the mole ratio 1/2 times 1.59 equals 0.796 moles of O2 V = nRT/P P=0.986 atm V is not known n=0.797222 moles of O2 R=0.0821 T=294 K (0.986 atm)V=(0.797222 moles of O2)(0.0821)(294 K) V=19.5 Liters of O2

OpenStudy (tywower):

is this correct? @Abhisar

OpenStudy (abhisar):

\(\huge \checkmark\)

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