2NO(g) + O2 (g) --> 2NO2 How many moles of O2 are required to produce 92.0 g NO2?

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$2NO(g) + O_2 (g) -> 2NO_2$

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Answer 1 · 2018-04-28T16:16:19

$2NO(g) + O_2 (g) -> 2NO_2(g)$

This equation tells us that for every $2$ moles of $NO_2$ formed ; we require $1$ mole of $O_2$ .

So given amount of $NO_2$ formed $= 92g$

Molar mass of $NO_2 = (14+16xx2)g=46g$

So number of moles of $NO_2$ formed $=92/46=2$ moles

$:.$ The mole of $O_2$ required here $=1$ mole