Question

What is the mass in grams of NO that will be produced from 30.0 g of NO₂ reacted with excess water in the following chemical reaction? 3 NO₂(g) + H₂O(l) → 2 HNO₃(g) + NO(g)

Answer 1

Mass of NO produced is "6.5 g".

Step-by-step explanation:

The given reaction is:

⇒
`3 NO_2(g) + H_2O(l) \rightarrow 2 HNO_3(g) + NO(g)`

Now,

⇒
`Moles \ of \ NO_2= (Mass \ of \ NO_2)/(Molar \ mass)`

⇒
`=(30 \ g)/(46 \ g \ mol^(-1))`

⇒
`=0.65 \ mol`

• We shouldn't have to acknowledge the sum of H₂O as it would be in excess. It's going to mot finish resolving answer.
• We provided 0.65 mol of NO₂, of which 3 mol of NO₂ = 1 mol of NO was previously given.

Accordingly,

`0.65 \ mol \ NO_2 = (1)/(3) * 0.65 \ mol \ NO` is created.

So,

The mass of NO will be:

=
`mol* molar \ mass`

=
`(1)/(3)* 65 \ mol* 30 \ g \ mol^(-1)`

=
`6.5 \ g`