How many liters of oxygen gas, at 285 k and 1.2 atm, will react with 35.4 grams of calcium metal? show all of the work used to solve this problem.

1) Chemical equation:

2Ca + O2 ---> 2CaO

2) molar ratios

2 mol Ca : 1 mol O2 : 2 mol CaO

3) Convert 35.4 g of Ca to moles:

number of moles = mass in grams / atomic mass = 35.4 g / 40.078 g/mol = 0.883 mol

3) use proportion and solve for x:

2 mol Ca / 1 mol O2 = 0.883 mol Ca / x

=> x = 0.883 mol Ca * 1 mol O2 / 2 mol Ca

=> x = 0.4416 mol O2

3) use ideal gases equation to transform moles to liters

pV = nRT => V = nRT / p =

V = [0.4416 mol * 0.0821 atm*liter / atm*K * 285K] / 1.2 atm = 8.6 liter

8.6 liters

Ur e n d is not real