When a sample of NO2(g) (849.9 grams) is placed in 390.0 L reaction vessel at 961.0 K and allowed to come to equilibrium the mixture contains 248.2 grams of NO(g). What is the concentration (mol/L) of NO2(g)?

Question

When a sample of NO2(g) (849.9 grams) is placed in 390.0 L reaction vessel at 961.0 K and allowed to come to equilibrium the mixture contains 248.2 grams of NO(g). What is the concentration (mol/L) of NO2(g)?

2NO2(g) = 2NO(g)+O2(g)

I keep getting .0212 or .04736. This is what I'm doing [849.9/molar mass of no2]/83.00L but its not the right answer. The right answer is 0.02616.

Thanks

Answer 1

I assume you want the NO2 at equilibrium although you don't say that in the post.

initial:
NO2 = 849.9/(46*390) = ?? moles/L
NO = 0
O2 = 0

equilibrium:
O2 = not needed.
NO = 248.2/(30*390) = ?? moles/L
NO2 = -----
Since 2 mols NO are produce for each 2 moles NO2 initially, those molecules are 1:1 so subtract NO2 initial - NO equilibrium = NO2 equilibrium.
I did the math and have
NO2 initial = 0.04737 M
NO equilibrium = 0.02121 M
0.04737-0.02121 = 0.02616 M.
:-)

Answer 2

Thanks Dr.Bob

Answer 3

To find the concentration of NO2(g) in mol/L, we need to use the information given in the question and the equation provided.

First, let's calculate the number of moles of NO(g) present in the reaction vessel. We can do this by dividing the given mass of NO(g) (248.2 grams) by its molar mass.

The molar mass of NO(g) is calculated as follows:
Molar mass of N = 14.01 g/mol
Molar mass of O = 16.00 g/mol

So, the molar mass of NO(g) = (14.01 g/mol) + (16.00 g/mol) = 30.01 g/mol

The number of moles of NO(g) = 248.2 g / 30.01 g/mol ≈ 8.270 mol

According to the balanced equation, 2 moles of NO2(g) are produced for every 2 moles of NO(g) consumed. Therefore, the number of moles of NO2(g) is also 8.270 mol.

Now, let's calculate the volume in liters. We know that 849.9 grams of NO2(g) is placed in a 390.0 L reaction vessel.

To convert grams to moles of NO2(g), we divide the given mass by its molar mass:
The molar mass of NO2(g) can be calculated as follows:
(14.01 g/mol) + (2 * 16.00 g/mol) = 46.01 g/mol

Therefore, the number of moles of NO2(g) = mass / molar mass = 849.9 g / 46.01 g/mol ≈ 18.473 mol

Finally, we can calculate the concentration (mol/L) of NO2(g) by dividing the number of moles of NO2(g) by the volume of the reaction vessel:

Concentration of NO2(g) = moles of NO2(g) / volume of reaction vessel
= 18.473 mol / 390.0 L ≈ 0.04736 mol/L

From your calculation, it seems that you used a different molar mass for NO2(g). Please make sure to use the correct molar masses for both NO(g) and NO2(g) in your calculations.

The correct answer for the concentration of NO2(g) is approximately 0.04736 mol/L or 0.0464 mol/L, rounded to four significant figures. The answer provided in the question (0.02616 mol/L) appears to be incorrect unless there is a mistake in the given values or equation.