Answer:
You will have 19.9L of Cl2
Step-by-step explanation:
We can solve this question using:
PV = nRT; V = nRT/P
Where V is the volume of the gas
n the moles of Cl2
R is gas constant = 0.082atmL/molK
T is 273.15K assuming STP conditions
P is 1atm at STP
The moles of 63g of Cl2 gas are -molar mass: 70.906g/mol:
63g * (1mol / 70.906g) = 0.8885 moles
Replacing:
V = 0.8885mol*0.082atmL/molK*273.15K/1atm
V = You will have 19.9L of Cl2