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?

Md Ariful Islam

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Md Ariful Islam

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