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.

I balanced the equation 2N2+2O2 = 2N2O5

BUT I AM NOT GETTING THE CORRECT ANSWER, WHAT AM I DOING WRONG?

you balanced the equaton incorrectly.

Based on the balanced equation you provided, the stoichiometry of the reaction is 2 moles of oxygen gas (O2) react to produce 2 moles of nitrogen pentoxide gas (N2O5). However, when calculating the volume of oxygen gas that reacts, we need to take into account the reaction conditions.

To solve this problem, we can use the ideal gas law, which states that PV = nRT, where P is pressure, V is volume, n is the number of moles, R is the gas constant, and T is temperature.

First, let's convert the given volume of N2O5 into moles. We need to find the number of moles of N2O5 that corresponds to 750.0 cm3. We can use the relationship between volume, pressure, and temperature:

PV = nRT

Since we have the pressure (2.50 atm), volume (750.0 cm3), and temperature (350°C), we can rearrange the equation to solve for moles:

n = PV / RT

Remember to convert the temperature to Kelvin (K) by adding 273.15:

n = (2.50 atm) x (750.0 cm3) / [(0.0821 L.atm/mol.K) x (350 + 273.15 K)]

Now, we know the number of moles of N2O5 that reacted. Since the stoichiometry of the balanced equation is 2 moles of O2 reacting to produce 2 moles of N2O5, we can conclude that 2 moles of O2 react to yield 750.0 cm3 of N2O5.

Therefore, the volume of oxygen gas that reacts is also 750.0 cm3.

If your answer is not matching the correct answer, double-check your conversions and ensure that you used the correct values for pressure, volume, temperature, and the gas constant (0.0821 L.atm/mol.K).