N2O5 is an unstable gas that decomposes according to the following reaction:

2N2O3(g) ---> 4NO2(g) + O2(g)
What would be the total pressure of gases present if a 10.0L container at 22.0 degree centigrade begins with 0.400 atm of N2O5 and the gas completely decomposes?

You made a typo on the equation.

2N2O5 ==> 4NO2 + O2
When using gases one may use the gas pressure as if it were mols.
Convert atm N2O5 to atm NO2. That's 0.4 atom x (4 mols NO2/2 mols N2O5) = ? atm for NO2.
0.4 atm x (1 mol O2/2 mols N2O5) - ? atm for O2.
Add the two for Ptotal.

1 atm

Well, if N2O5 completely decomposes according to the given reaction, then all of the initial N2O5 will turn into products, namely NO2 and O2. So, we need to find the amount of moles of N2O5 and then use stoichiometry to calculate the moles of NO2 and O2.

Using the ideal gas law, we can find the moles of N2O5:
PV = nRT
(0.400 atm)(10.0 L) = n(0.0821 L·atm/mol·K)(295 K)
Simplifying this equation, we find that n (moles of N2O5) = 0.018 moles.

According to the stoichiometry of the reaction, 1 mole of N2O5 produces 4 moles of NO2 and 1 mole of O2. Therefore, we have:
0.018 moles N2O5 x (4 moles NO2 / 2 moles N2O5) = 0.036 moles NO2
and
0.018 moles N2O5 x (1 mole O2 / 2 moles N2O5) = 0.009 moles O2.

Now, to find the total pressure of gases present, we can use the ideal gas law again:
PV = nRT
We have two different gases, so we'll need to calculate the partial pressure for each gas and then add them up to get the total pressure.

Partial pressure of NO2:
(0.036 moles)(0.0821 L·atm/mol·K)(295 K) / 10.0 L = 0.086 atm

Partial pressure of O2:
(0.009 moles)(0.0821 L·atm/mol·K)(295 K) / 10.0 L = 0.020 atm

Therefore, the total pressure of gases present is:
0.086 atm + 0.020 atm = 0.106 atm.

So, the total pressure of gases present in the container would be approximately 0.106 atm. But enough with the math, let's just say the pressure is "a bunch of gas particles having a grand old time together." 🎈😄

To find the total pressure of gases present after the decomposition of N2O5, we need to calculate the partial pressures of each gas using the ideal gas law.

The ideal gas law is expressed as:

PV = nRT

where:
P is the pressure of the gas,
V is the volume of the gas,
n is the number of moles of the gas,
R is the ideal gas constant (0.0821 L·atm/(mol·K)),
T is the temperature in Kelvin.

Given that the container has a volume of 10.0 L and a temperature of 22.0 degrees Celsius (which needs to be converted to Kelvin), we can proceed with the calculation.

Step 1: Convert the temperature to Kelvin
T(K) = T(°C) + 273.15
T(K) = 22.0 + 273.15
T(K) = 295.15 K

Step 2: Calculate the initial number of moles of N2O5
n(N2O5) = PV / RT
n(N2O5) = (0.400 atm)(10.0 L) / (0.0821 L·atm/(mol·K))(295.15 K)
n(N2O5) ≈ 0.171 mol

Step 3: Apply the stoichiometry of the decomposition reaction
For every 2 moles of N2O5 decomposed, we get 4 moles of NO2 and 1 mole of O2.

Since the initial number of moles of N2O5 is 0.171 mol, the final number of moles of NO2 will be 4 times that value, and the final number of moles of O2 will be the same as the initial number of moles of N2O5.

n(NO2) = 4 * n(N2O5)
n(NO2) = 4 * 0.171 mol
n(NO2) ≈ 0.684 mol

n(O2) = n(N2O5)
n(O2) ≈ 0.171 mol

Step 4: Calculate the partial pressures of each gas
P(NO2) = n(NO2) * RT / V
P(NO2) = (0.684 mol)(0.0821 L·atm/(mol·K))(295.15 K) / 10.0 L
P(NO2) ≈ 1.599 atm

P(O2) = n(O2) * RT / V
P(O2) = (0.171 mol)(0.0821 L·atm/(mol·K))(295.15 K) / 10.0 L
P(O2) ≈ 0.379 atm

Step 5: Calculate the total pressure
The total pressure is the sum of the partial pressures of each gas.
Total pressure = P(NO2) + P(O2)
Total pressure = 1.599 atm + 0.379 atm
Total pressure ≈ 1.978 atm

Therefore, the total pressure of gases present after the decomposition of N2O5 in the 10.0L container at 22.0 degrees Celsius would be approximately 1.978 atm.