Question

Consider the following reaction:
2NiO (s)--> 2Ni(s) + 2O2 (g)

If O2 is collected over water at 40.0 C and a total pressure of 752 mmHg, what volume of gas will be collected for the complete reaction of 25.67 g of NiO?

I am struggling to set up this problem. Can anyone help me get started or explain the steps needed to arrive at an answer? I know that PV=nRT but I don't know how to use gas laws to solve this. Thank you.

Answer 1

The reaction written correctly is:
2NiO (s)--> 2Ni(s) + O2 (g)
1. Divide 25.67g NiO by the formula its mass to get the number of moles of NiO.
2. Since 2 moles NiO produce 1 mole O2,
moles of O2 = ( ___mol NiO)(1/2)
3. Look up the vapor pressure of water, P(vapor), at 40.0 deg C.
4. The partial pressure of O2 is:
P(O2 = 752mmHg-P(vapor)
5. Divide the partial pressure of oxygen, P(O2), by 760mmHg/atm to convert the gas pressure to atmospheres.
6. Find the volume of O2 using PV=nRT
V = (unknown), P = PO2 in atm., n = moles of O2, R = 0.0821L.atm/K.mol, T = (40 + 273) degrees Kelvin

Answer 2

Thanks so much! I got 4.82 which seems to be correct. Thank you again