According to the following balanced reaction, how many moles of NO are formed from 2.58 moles of [tex]NO_2[/tex] if there is plenty of water present?

[tex]\[
3 NO_2(g) + H_2O(l) \rightarrow 2 HNO_3(aq) + NO(g)
\][/tex]

A. 7.50 moles NO

B. 3.75 moles NO

C. 2.50 moles NO

D. 1.67 moles NO

E. 0.834 moles NO



Answer :

To determine the number of moles of [tex]\( \text{NO} \)[/tex] formed from 2.58 moles of [tex]\( \text{NO}_2 \)[/tex] given the balanced chemical reaction:

[tex]\[ 3 \, \text{NO}_2(g) + \text{H}_2\text{O}(l) \rightarrow 2 \, \text{HNO}_3(aq) + \text{NO}(g) \][/tex]

follow these steps:

1. Identify the stoichiometric coefficients:
- From the balanced reaction, 3 moles of [tex]\( \text{NO}_2 \)[/tex] produce 1 mole of [tex]\( \text{NO} \)[/tex].

2. Write the mole ratio:
- The ratio of moles of [tex]\( \text{NO} \)[/tex] to moles of [tex]\( \text{NO}_2 \)[/tex] is [tex]\( \frac{1 \text{ mole NO}}{3 \text{ moles NO}_2} \)[/tex].

3. Calculate the moles of [tex]\( \text{NO} \)[/tex] produced:
- Using the mole ratio, you can find the moles of [tex]\( \text{NO} \)[/tex] produced from 2.58 moles of [tex]\( \text{NO}_2 \)[/tex].

[tex]\[ \text{moles of NO} = 2.58 \, \text{moles of} \, \text{NO}_2 \times \frac{1 \text{ mole of NO}}{3 \text{ moles of} \, \text{NO}_2} \][/tex]

4. Perform the multiplication:
- Calculate the result:

[tex]\[ \text{moles of NO} = 2.58 \times \frac{1}{3} \][/tex]

Therefore, the number of moles of [tex]\( \text{NO} \)[/tex] produced is [tex]\( 0.86 \)[/tex] moles.

So, from 2.58 moles of [tex]\( \text{NO}_2 \)[/tex] with plenty of water present, 0.86 moles of [tex]\( \text{NO} \)[/tex] will be formed.