The reaction between $H_2$ and $O_2$ produces 13.1 g of water. How many grams of $O_2$ reacted? $2H_2(g) + O_2(g) -> 2H_2O(g)$?

Md Ariful Islam

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Md Ariful Islam

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There are $(13.1*g)/(18.01*g*mol^-1)=0.727*mol$ of water product.

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

Given the of the reaction, there were necessarily $0.727*mol$ dihydrogen reactant, and $(0.727*mol)/2$ dioxygen reactant. I speak of dihydrogen, dioxygen etc. because I wish to avoid the confusion of mistaking atoms for molecules and vice versa.

$"Mass of dioxygen reactant "=(0.727*mol)/2xx32.00*g*mol^-1=??g$

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