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

Answer : 9.89 L

Explanation;

The balanced reaction equation is 
2Ca(s) + O₂(g)→ 2CaO(s)

Moles (mol) = Mass (g) /Molar mass (g/mol)

Mass of Ca = 35.4 g
Molar mass of Ca = 40.08 g/mol
Hence, moles of Ca = 35.4 g /40.08 g/mol
                                = 0.883 mol

The stoichiometric ratio between Ca and O₂ is 2 : 1.

Hence, 
   reacted moles of O₂ = moles of Ca / 2
                                      = 0.883 mol / 2
                                      = 0.4415 mol

Let's assume that O₂ is an ideal gas.

1 mole of ideal gas has the volume as 22.4 L at STP (273 K , 1 atm)

Hence,
   volume of 0.4415 mol of O₂ = 22.4 L mol⁻¹ x 0.4415 mol
                                                  = 9.8896 L
                                                  = 9.89 L 

What’s the solution?

Ametal with a large number of free flowing electrons most likely has good conductivity