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 equation incorrectly. You have 4 O on the left, 10 on the right.
From the balanced equation, we can see that the stoichiometric ratio between O2 and N2O5 is 2:2, or 1:1. This means that for every 1 mole of N2O5 formed, we need 1 mole of O2.
To calculate the volume of oxygen gas required, we need to convert the given volume of N2O5 to moles, then use the stoichiometric ratio to determine the moles of O2, and finally convert the moles of O2 to volume using the ideal gas law.
1. Convert the given volume of N2O5 to moles:
Using the ideal gas law equation PV = nRT, we can rearrange it to solve for moles: n = PV / RT.
n(N2O5) = (750.0 cm3) / (0.0821 L·atm/mol·K * 350°C + 273.15 K).
2. Determine the moles of O2:
Since the stoichiometric ratio between N2O5 and O2 is 1:1, the moles of O2 required is equal to the moles of N2O5 obtained in step 1.
3. Convert the moles of O2 to volume:
Using the ideal gas law equation V = nRT / P, we can rearrange it to solve for volume: V = (n(O2) * R * T) / P.
V(O2) = (n(O2) * 0.0821 L·atm/mol·K * 350°C + 273.15 K) / 2.50 atm.
Make sure to check that all unit conversions were done correctly. Double-check the values of R (the ideal gas constant) and the temperature (in Kelvin) used in the calculations.
By following these steps, you should be able to calculate the correct volume of oxygen gas required.