Question

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

Answer 1

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`.