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

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Chemistry

Elsa Mccullough

1 June, 18:18

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

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Jax Bright · 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