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  4. The moles of O2 required for reacting with 6.8g ammonia(...NH3 + .... O2---->....NO +....H2O) is

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Q. The moles of $O_2$ required for reacting with 6.8g ammonia$(...NH_3 + .... O_2---->....NO +....H_2O)$ is

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Solution:

The balanced equation is
$4NH_{3}+5O_{2}\to 4NO+6H_{2}O$
From the equation ;
$4\times 17\,g$ of $NH_{3}$ require $O_{2}=5\,mol$
$6.8\,g$ of $NH_{3}$ require $O_{2}=\frac{5\times8.5}{4\times17}$
$=0.5\,mol$