Question

PLEASE HELP REALLY QUICK!!

How many liters of chlorine gas can react with 56.0 grams of calcium metal at standard temperature and pressure?
Ca + Cl2 ----> CaCl2

How go I go about solving this? Where do I even start?

Answer 1

. . . . . Ca . . . + . . . Cl2 . . . ----------> . . . . CaCl2

. . . . 1 mole . . . . 1 mole . . . . . . . . . . . . . 1 mole


n(Ca) = m(Ca) / M(Ca) = 56 / 40,08 = 1,397 mole of Ca

According the reaction equation :

1 mole of Ca requires 1 mole of Cl2


... then 1.397 mole of Ca requires 1.397 mole of Cl2


The necessary volume of Cl2 is therefore :

V (Cl2) = Vm x n (Cl2) = 22.4 x 1,397 = 31,30 Litres of Cl2

Answer 2

Thanks so much, I noticed I posted it in the wrong section. Where do you get the 40 from ?

Answer 3

sorry for the late response. 40 is the mass of Ca

Answer 4

No worries, and the 22.4?

Answer 5

that is the stand temperature and pressure.
1mol=22.4 Liters at STP

Answer 6

Ahhh, okay. So the answer would be 31.30? not 31.36?

Answer 7

yep 31.30L Cl2 at STP