Please help me out? 10.0 grams of water would produce what volume?

Question

The electrolysis of 10.0 grams of water would produce what volumes in dm3 of O2 and H2 at STP? Give each electrode reaction.

Answer 1

At 1 atm press and 25 deg centigrade 1gm of water = 1ml of water = 1cc of water

So, 10 gms of water = 10 ml of water = 10 cm3

1 dm3 = 1000 cm3

so, 10cm3 = 0.01 dm3

H2+O2= 2H20

Answer 2

10.0 g H2O = 0.556 moles

2H2O --> 2H2(g) + O2(g)

You get 0.556 moles of H2 which is 12.4 L at STP since 1 mol = 22.4 L.  And you get half that volume of O2 because you get half as many moles of O2, or 6.2 L of O2.

This is the kind of problem I have my students do without a calculator, in their heads, while standing beside their desk.  Except I would told them that it was 9.0 grams of water, to make the mental math even easier.

Answer 3

Cathode:  4H+(aq) + 4e- ==>  2H2(g)

Anode :  2H2O(l) ==>  O2(g) + 4H+(aq) + 4e-

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Net rxn: 2H2O(l) ==>  2H2(g) + O2(g)

10.0 g H2O x (1 mole H2O / 18.0 g H2O)

x (2 moles H2 / 2 moles H2O) x (22.4 L H2 / 1 mole H2 at STP) = 12.4 L H2 = 12.4 dm^3 H2

10.0 g H2O x (1 mole H2O / 18.0 g H2O)

x (1 moles O2 / 2 moles H2O) x (22.4 L O2 / 1 mole O2 at STP) = 6.2 L O2 = 6.2 dm^3 O2