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The idea behind the question is that you know the partial pressure of N$_2$ is 0.96 atm (= 1 atm - 0.04 atm). And you know that the partial pressure of all the H$_2$ that can be produced from the water is thus (1.86 atm - 0.96 atm)* 2/3 = 0.6 atm. [Ignoring the volume of liquid water, which is only ~1/1000 that of the gas.] From this you can determine the ...


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I think the fact you are missing is that after combustion, water vapors will be given away and the volume lost during combustion will be equal to the amount of water lost. $\ce{KOH}$ will absorb $\ce{CO2}$ and the volume absorbed by $\ce{KOH}$ will be equal to volume of $\ce{CO2}$ in the solution. Thus, you will have two equations and you will be able to ...


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If you take the wrong amounts of each, you get not a pure product but some mixture with the educt! The law of constant proportions says exactly that: Whenever you do that reaction with hydrogen and oxygen properly, you always get a volume ratio of (hydrogen:oxygen:water gas) 2:1:2, and a mass ratio of 1:8:9.


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Your chemical equation is not balanced, you just dropped one oxygen atom! First balance it! It should be: $\ce{2 H2 + O2 -> 2 H2O}$ Then, you will get the correct masses.


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Your "should-be" solution, $$\ce{3H2O2 + 2NO3- + 2H+ → 3O2 + 2NO + 4H2O}$$ is a linear combination of $$\ce{H2O2 + 2NO3- + 2H+ → 2O2 + 2NO + 2H2O}$$ and $$\ce{2H2O2 → O2 + 2H2O}$$ Any other linear combination is valid as well.


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