You breathe in 12.0 l of pure oxygen at 298 k and 1,000 kpa to fill your lungs. how many moles of oxygen did you take in. can you also show me how to solve this ?

Your question must be at 1.000kPa not 1,000KPa... That's ~10Atms, or 10x atmospheric pressure we as humans are used to breathing... So the following calculation is for 1.000kPa or 0.987Atms. => 0.482 mole O₂(g). If your problem is at 1,000kPa then multiply mole answer by 10.

Explanation:

Using the Ideal Gas Law PV = nRT solve for n = PV/RT

n = moles of O₂(g)

P = Pressure = 1.000KPa = 0.987Atm

V = Volume in liters = 12L

R = Universal Gas Constant = 0.08206 L·Atm/mol·K

T = Temperature in Kelvin = 298K

n = P·V/R·T = (0.987Atm)(12L)/(0.08206 L·Atm/mol·K)(298K) = 0.482 mole O₂(g)

  the   balanced   equation   for   reaction of   ca metal   with   o2 to   produce   cao   is as below

2 ca+ o2→ 2 cao2

  explanation

the reaction is   balanced     since the   number of atoms   in reactant side is   equal to   number of atoms   in   product side.   for   example   they   are 2   atom of   ca in   reactant   side   and   2   atoms of ca   in   the product side.

  the   process   is a   meet the   requirement   of   redox   reaction because :     calcium   is   oxidized   from   oxidation state 0   to 2+   while   oxygen   is reduced from   oxidation   state   0   to 2-