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dell8606 · May 25, 2009, 12:18 PM

If 25g of hydrogen and 13g of oxygen ate allowed to react, how much water will be produced?

Perito · May 25, 2009, 12:31 PM

First identify the compounds or elements involved:

Hydrogen: H2
Oxygen: O2
Water H2O

Then write the equation

$H_2 + O_2 \rightarrow H_2O$

Then balance the equation.

$2H_2 + O_2 \rightarrow 2H_2O$

Calculate the molecular weights for the materials involved:

H2 - Molecular weight ≈ 2
O2 - Molecular weight ≈ 32
H2O - Molecular weight ≈ 18

Now figure out the number of moles involved:

$25\,g\,H_2\,\times\ \frac {1\,mole\,H_2}{2\,g\,H_2} = 12.5\,moles\,of\,H2$

$13\,g\,O_2\,\times\ \frac {1\,mole\,O_2}{32\,g\,O_2} \approx 0.41\,moles\,of\,O2$

According to the balanced equation, you need 2 moles of H2 for every mole of O2. You have 12.5 moles of H2, so this could react with 6.25 moles of O2. But, you have nowhere near that much O2. You only have 0.406 moles. Therefore, the most H2 that can react is 0.813 moles of H2. The rest of the H2 will sit there and do what unreacting molecules do (probably spend time with the senator from my state -- doing nothing).

So, you have 0.406 moles of O2 + 0.813 moles of H2 producing 0.813 moles of H2O (all this comes from the balanced equation).

$0.813\,moles\,H_2O\,\times\,18\,\frac {grams\,H_2O}{mole\,H_2O}\,\approx\,14.6\,g\,H_2O$

If you're more accurate than I was with molecular weights, you can get more accurate answers.