2K+2Hsub(2)O -> 2KOH + Hsub(2)?

If I start with 9 grams of Hsub(2)O, how many moles of Hsub(2) will be produced? Isn't it 1 mol of Hsub(2)?

Answer:
2K(s) + 2H2O(l) ---> 2KOH(aq) + H2(g)

We see that the mole ratio between H2O and H2 is 2:1

Ok, so we find the number of moles of water in 9g:

n = m/Mr (Mr = molecular mass of water = 18)
n = 9/18
n = 0.5 moles

Remember the mole ratio? so half as much in moles of H2 is produced:

0.5/2 = 0.25 moles of H2 (which is 0.5g of H2)
2K + 2H2O ---> 2KOH + H2
9 grams of H2O is 0.50 mole (9/18)
If 2 mole of H2O yield 1 mole of H2 (from the balanced equation), then 0.50 mole of H2O will yield 0.25 mole of H2.
From the stoichiometric equation, you know that 2 moles of water (H2O) gives 1 mole of hydrogen (H2).

Now work out the Mr of both molecules:

Water = H*2 + O = 1*2 + 16 = 18
Hydrogen = H*2 = 1*2 = 2

So 18g of water should give (2/2) = 1g of hydrogen
9g of water should give 1/2 = 0.5g of hydrogen.

0.5g of hydrogen is 0.25 moles (1/4).

Hope this helps!

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