In the combustion of hydrogen gas, hydrogen reacts with oxygen from the air to form water vapor. Hydrogen + Oxygen $->$ water. If you burn 58.1 g of hydrogen and produce 519 g of water how much oxygen reacted?

Md Ariful Islam

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The stoichiometrically balanced equation tells us unequivocally that $1$ $mol$ dihydrogen reacts with $0.50$ $mol$ dioxygen to give $1$ $mol$ water.

$"Moles of dihydrogen"$ $=$ $(58.1*g)/(2.02*g*mol)$ $=$ $28.7*mol$ $H_2$

$"Moles of water"$ $=$ $(519.0*g)/(18.02*g*mol)$ $=$ $28.7*mol$ $H_2O$

The water is stoichiometric with respect to the amount of dihydrogen. (What do I mean by this?) There was thus a stoichiometric quantity of dioxygen gas, i.e. $14.4*mol$ $O_2$

$"Mass of dioxygen"$ $=$ $14.4*cancel(mol)xx32.00*g*cancel(mol^-1)$ $=$ $??g$

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