How many liters of oxygen gas, at standard temperature and pressure, will react with 35.4 grams of calcium metal? show all of the work used to solve this problem.   2ca + o2 > 2cao pls < 3

At STP, P = 1 atm, and T = 0 C
Thus, PV = nRT => V = nR(273). We will use this later...
if you have 35.4 Ca, and the molar mass of Ca is 40.08, you get .883 moles Ca. Thus, since it takes 2 moles of Ca to form a reaction, you only need half the moles of Ca of O2. Thus, n(O2) = .883/2
Tie this back to the first equation and you get
V = .442 * 0.082057(which is R) * 273 = 9.9 L