When 1.57 mol O_2 reacts with H_2 to form H_2O, how many moles of H_2 are consumed in the process?

1 Answer
anor277
Dec 9, 2016

Look at the stoichiometry, the chemical equivalence, of the reaction:

H_2(g) + 1/2O_2(g) rarr H_2O(l)

Explanation:

H_2(g) + 1/2O_2(g) rarr H_2O(l), or I could simply double the equation,

2H_2(g) + O_2(g) rarr 2H_2O(l)

Now, clearly, for each mole/equiv of dioxygen, 2 moles/equivs dihydrogen react. If there are 1.57*mol O_2, stoichiometric equivalence requires 2xx1.57*mol dihydrogen. (i) How many moles of water will be generated; and (ii) what is the mass of the water?

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