The solution is as follows:
Always start with the basic equation:
H2 + O2 -----> H2O(l)
Balancing the equation above gives:
2H2 + O2------> 2H2O(l)
Since 200 molecules of H2 & O2 are involved,just multiply the equation above with 200,
200(2H2 + O2)---->200(2H2O)
400H2 + 200 O2------> 400 H2O(l)
The answer is 400 water molecules
Hope my explanation helps :)
I disagree with the answer of 400 molecules H2O produced. Only 200 will be produced.
This is a limiting reagent problem. I work these by calculating, with two separate calculations, the amount of H2O that can be formed, then using the smaller value. For example,
If we take 200 molecules H2 and all of the O2 we need, we could produce
200 H2 molecules x (2 mols H2O/2 mol H2) = 200 molecules H2O.
If we take 200 molecules O2 and all of the H2 we need, we could produce
200 O2 molecules x (2 moles H2O/1 mol O2) = 400 molecules H2O.
So you get 200 for one answer; 400 for another answer and that means one of them is wrong. The correct value in limiting reagent problems is ALWAYS the smaller value and the reagent producing that number is the limiting reagent. Therefore, you can produce 200 molecules H2O and you will have some O2 remain unreacted. You can calculate how much O2 remains this way.
200 molecules H2 x (1 mol O2/2 mols H2) = 100 molecules O2 used in the reaction
Since you had 200 molecules O2 initially and used 100, you must have 100 O2 molecules that didn't react.
Thank you Dr.Bob, now I know my mistake.