Consider the formation of nitrogen dioxide from nitric
oxideand oxygen.
2NO(g) + O2(g) -- 2NO2(g)
If 9L of NO are reacted with excess 02 at STP, what is
thevolume in liters of the NO2 produced?

Respuesta :

Answer:

9 L

Explanation:

According to the question , the given reaction is -

2NO(g) + O₂(g)------->2NO₂(g)

Since ,

At STP ,

One mole of a gas occupies the volume of 22.4 L.

Hence , as given in the question -

9 L of NO , i.e .

22.4 L = 1 mol

1 L = 1 / 22.4 mol

9 L = 1 / 22.4  * 9 L = 0.40 mol

From the chemical reaction ,

The Oxygen is in excess , hence NO becomes the limiting reagent , and will determine the moles of product .

Hence ,  

2 moles of NO will produce 2 moles of NO₂.

Therefore ,

0.40 mol of NO will produce 0.40 mol of NO₂.

Hence , the volume of NO₂ can be calculated as -

1 mol = 22.4 L

0.40 mol = 0.40 * 22.4 L = 9 L