Given that 4 NH3+5 O2 --> 4 NO + 6H2O, when 4.50 mol of H2O are formed, the amount of NO formed is?

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Given that 4 NH3+5 O2 --> 4 NO + 6H2O, when 4.50 mol of H2O are formed, the amount of NO formed is?

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Answer 1 · 2018-04-08T17:24:57

The given balanced equation is,

$4 NH_3+5 O_2 -> 4 NO + 6H_2O$

This implies, $4 " moles"$ of $NH_3$ reacts with $5 " moles"$ of $O_2$ to produce $4 " moles"$ of $NO$ and $6 " moles"$ of $H_2O$

So, the net products will be, $4 " moles"$ of $NO$ and $6 " moles"$ of $H_2O$

Accordingly, $x " moles"$ of $NO$ and $4.5 " moles"$ of $H_2O$ should be formed.

where $x$ would be,
$x=(4xx4.5)/6$

$x=3 " moles"$

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Note : The $4:6=> 2:3$ ratio remains constant for the given pair of products.
As from the question, if the answer $3 " moles"$ is correct, then $3:4.5$ must be equal to $2:3$

Checking that, $3:9/2 => 6:9 => 2:3$

therefore, the answer must be correct.

$3 " moles"$ of $NO$ is formed along with $4.50 " moles"$ of $H_2O$ .

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Given that 4 NH3+5 O2 --> 4 NO + 6H2O, when 4.50 mol of H2O are formed, the amount of NO formed is?

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