Question

How many liters of oxygen gas, at standard temperature and pressure, will react with 35.4 grams of calcium metal? No idea how to solve this. :/

Answer 1

2 Ca . . . + . . . O2 . . . -----> . . . 2 CaO
. . . . . 40,08 g/mol


n(Ca) = m(Ca) / M(Ca)
n(Ca) = 35,4 / 40,08
n(Ca) = 0,8832 mol of Ca


2 mol of Ca requires 1 mol of O2

. . . then :

0,8832 mol of Ca requires 0,8832 / 2 = 0,4416 mol of O2


V(O2) = Vm x n(O2)
V(O2) = 22,4 x 0,4416
V(O2) = 9,89 L of O2