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

Answer:

Mass of NO produced is "6.5 g".

Explanation:

The given reaction is:

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

Now,

⇒ $Moles \ of \ NO_2= \frac{Mass \ of \ NO_2}{Molar \ mass}$

⇒                          $=\frac{30 \ g}{46 \ g \ mol^{-1}}$

⇒                          $=0.65 \ mol$

• 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,  

$0.65 \ mol \ NO_2 = \frac{1}{3} \times 0.65 \ mol \ NO$  is created.

So,

The mass of NO will be:

= $mol\times molar \ mass$

= $\frac{1}{3}\times 65 \ mol\times 30 \ g \ mol^{-1}$

= $6.5 \ g$