9.86 x 10²⁸ O-atoms require what volume (L) N₂O₂ at STP?

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Maximus Mooney

Chemistry

15 September, 12:55

9.86 x 10²⁸ O-atoms require what volume (L) N₂O₂ at STP?

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Serenity Blevins · Answer 1

First, we use avogadro's number to convert atoms into moles. Then, relate the number of moles from elemental to the compound. Lastly, we use conditions at STP to calculate the volume. We do as follows:

9.86 x 10²⁸ O-atoms (1 mol / 6.022x10^23 atoms O) (1 mol N2O2 / 2 mol O) (22.4 L / 1 mol) = 1833809.37 L needed