In the formation of smog, nitrogen and oxygen gas react to form nitrogen dioxide:
N2(g)+2O2(g)→2NO2(g)How many grams of NO2 will be produced when 2.1L of nitrogen at 840mmHg and 28∘C are completely reacted?
I know that you use the equation n=PV/RT to get the number of moles of nitrogen. What I am stuck on is getting the moles of NO2
Look at the coefficients in the balanced equation. With those you can convert mols of anything you HAVE in the equation to anything you WANT in the equation.
mols N2 x (2 mol NO2/1 mol N2) = mols N2 x 2/1 = ? mols NO2.
When I solved the equation I got 3.1 grams of NO2. That isn't the correct answer. What did I do wrong?
I don't know. My crystal ball is cloudy tonight. :-)
If you will post your work I will find the error.
now you just need to multiply 3.1 by two= 6.2 is the answer... because there two mole of NO2 on one mole of N2
Well, it looks like the nitrogen is just dying to react with oxygen and form some NO2, huh? Let's see if we can calculate the moles of NO2 that will be produced and make this reaction a gas!
First, we'll need to find the number of moles of nitrogen using the ideal gas law equation you mentioned. Let's plug in the given values.
P = 840 mmHg
V = 2.1 L
T = 28 degrees Celsius = 28 + 273 = 301 K
R = the gas constant = 0.0821 L·atm/(mol·K)
Using n = PV/RT, we get:
n(N2) = (840 mmHg * 2.1 L) / (0.0821 L·atm/(mol·K) * 301 K)
Now, let's calculate n(N2):
n(N2) = (1764 mmHg·L) / (24.7381 L·atm·K⁻¹·mol⁻¹)
n(N2) ≈ 71.37 mol (approximately)
Now that we have the number of moles of nitrogen, we can determine the moles of NO2 produced by referring to the balanced equation.
According to the equation:
1 mol N2 produces 2 mol NO2
So, if we have 71.37 mol of N2, we can calculate the moles of NO2 produced as follows:
n(NO2) = 2 * n(N2)
n(NO2) = 2 * 71.37 mol
Therefore,
n(NO2) ≈ 142.74 mol
So, approximately 142.74 moles of NO2 will be produced when 2.1 L of nitrogen at 840 mmHg and 28 °C are completely reacted. Remember, though, that we need to convert this to grams.
I hope these calculations didn't take the fun out of the reaction for you!
To find the moles of NO2 produced, you need to use the stoichiometry of the balanced equation. The coefficients in the balanced equation give you the mole ratio between compounds.
Here's how you can calculate the moles of NO2 produced:
1. Convert the volume of nitrogen gas (2.1L) to moles using the ideal gas law equation, n = PV/RT. Plug in the values:
- P = 840 mmHg (convert to atm by dividing by 760)
- V = 2.1 L
- R = 0.0821 L.atm/mol.K (gas constant)
- T = 28°C + 273.15 = 301.15 K (convert to Kelvin)
Substitute these values into the ideal gas law equation and solve for n (moles of nitrogen gas).
2. Use the balanced equation to determine the mole ratio between nitrogen and nitrogen dioxide. From the balanced equation:
- 1 mole of N2 produces 2 moles of NO2.
- This means for every mole of N2 reacted, 2 moles of NO2 are produced.
3. Multiply the moles of N2 (obtained in step 1) by the mole ratio from the balanced equation to find the moles of NO2 produced.
4. Finally, to find the grams of NO2 produced, multiply the moles of NO2 by the molar mass of NO2 (46.0 g/mol).
Let's go through the calculations step-by-step:
Step 1:
Using the ideal gas law equation: n = PV/RT
P = 840 mmHg / 760 mmHg/atm = 1.105 atm
V = 2.1 L
R = 0.0821 L.atm/mol.K (gas constant)
T = 301.15 K
n = (1.105 atm * 2.1 L) / (0.0821 L.atm/mol.K * 301.15 K)
Solve for n to get the moles of nitrogen gas (N2).
Step 2:
From the balanced equation: 1 mole of N2 produces 2 moles of NO2.
Step 3:
Multiply the moles of N2 obtained in step 1 by the mole ratio from the balanced equation.
Step 4:
Multiply the moles of NO2 by the molar mass of NO2 (46.0 g/mol) to get the grams of NO2 produced.
That's how you can calculate the moles of NO2 and convert it to grams using the given information and the balanced equation.