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

Question

$2 N O (g) + O_{2} (g) \to 2 N O_{2}$ $2NO(g) + O_2 (g) -> 2NO_2$

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Answer 1 · 2018-04-28T16:16:19

$2 N O (g) + O_{2} (g) \to 2 N O_{2} (g)$ $2NO(g) + O_2 (g) -> 2NO_2(g)$

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

So given amount of $N O_{2}$ $NO_2$ formed $= 92 g$ $= 92g$

Molar mass of $N O_{2} = (14 + 16 \times 2) g = 46 g$ $NO_2 = (14+16xx2)g=46g$

So number of moles of $N O_{2}$ $NO_2$ formed $= \frac{92}{46} = 2$ $=92/46=2$ moles

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