N2(g) + O2(g) → N2O5(g)

What volume of oxygen gas reacts to yield 750.0 cm3 of N2O5? The reaction conditions are 350°C and 2.50 atm pressure

See your SO2/SO3 question below. Same process.

To find the volume of oxygen gas required, we need to first use the ideal gas law equation, PV = nRT, where P is the pressure, V is the volume, n is the number of moles, R is the ideal gas constant, and T is the temperature in Kelvin.

First, we need to convert the given volume from cm3 to liters, since R is usually given in liters.

1 L = 1000 cm3

Therefore, the volume of N2O5 is 750.0 cm3 ÷ 1000 = 0.750 L.

Next, we need to convert the temperature from Celsius to Kelvin.

T(K) = T(°C) + 273.15

T(K) = 350 + 273.15 = 623.15 K

Now, we need to rearrange the ideal gas law equation to solve for the number of moles of the reactants.

n = PV / RT

n(O2 gas) = (P(O2 gas) * V(N2O5)) / (R * T)

Substituting the given values:
P(O2 gas) = 2.50 atm
V(N2O5) = 0.750 L
R = 0.0821 L.atm/(mol.K)
T = 623.15 K

n(O2 gas) = (2.50 atm * 0.750 L) / (0.0821 L.atm/(mol.K) * 623.15 K)

Calculating the number of moles of oxygen gas:
n(O2 gas) ≈ 0.2276 mol

Finally, we can use the stoichiometric coefficients from the balanced equation to relate the number of moles of oxygen gas to the number of moles of N2O5. From the equation:

N2(g) + O2(g) → N2O5(g)

We can see that the molar ratio of O2 gas to N2O5 is 1:1.

Therefore, the number of moles of oxygen gas consumed is equal to the number of moles of N2O5 formed.

So, the volume of oxygen gas that reacts to yield 750.0 cm3 of N2O5 is 0.2276 mol (or 0.2276 liters).