Question

If 9.0 L of NO are reacted with excess O2 at STP, what is the volume in liters of the NO2 produced? Methane, the principal component of natural gas, is used for heating and cooking. The combustion process is CHA (g) + 20 (g) > CO2 (8) + 2H20(1)

Answer 1

9 L of the NO2 was produced if 9.0 L of NO are reacted with excess $O_2$ at STP.

What is an ideal gas equation?

The ideal gas law (PV = nRT) relates the macroscopic properties of ideal gases. An ideal gas is a gas in which the particles (a) do not attract or repel one another and (b) take up no space (have no volume).

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.

As given in the question -

9 L of NO, i.e .

22.4 L = 1 mol

1 L = $\frac{1}{22.4 mol}$

For 9 L =$\frac{1 X 9L}{22.4 mol}$ = 0.40 mol

From the chemical reaction,

The Oxygen is in excess, hence NO becomes the limiting reagent, and will determine the moles of the 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 X 22.4 L = 9 L

Hence, 9 L of the NO₂ was produced.

Learn more about the ideal gas here:

brainly.com/question/27691721

#SPJ1