Question

How many liters of oxygen gas, at 285 K and 1.2 atm, will react with 35.4 grams of calcium metal?

Answer 1

This look like some ideal gas calculation.
2 Ca + O2 -> 2 CaO
V=nRT/P

V: what we want to know
T: known constant
R: gas constant
P: known constant.
n: unknown

n(O2)=½*n(Ca)=½*35.4 g / M(Ca)

Answer 2

oh ok I wasnt sure if it was more stoichiometry or gas calculations, thanks a lot

Answer 3

Well both as we can call it gas stoichiometry :)