Question

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 HELP <3

Answer 1

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