PV = nRT
P is pressure, V is volume, n is number of moles, R is the gas constant, T is temperature in K
(2.85 atm)(12.5 L) = (n)(.08206)(27 C + 273)
n = 1.45 moles x 35.45 grams / mol Cl2 = 51.3 grams
Answer: The mass of bromine gas contained is 101.15 grams.
Explanation:
To calculate the mass of bromine gas, we use the ideal gas equation, which is:
PV = nRT
where,
P = pressure of the gas = 0.986 atm
V = Volume of the gas = 15.7 L
n = Number of moles of gas = ? mol
R = Gas constant = [tex]0.0821\text{ L atm }mol^{-1}K^{-1}[/tex]
T = temperature of the gas = [tex]24.6^oC=(273+24.6)K=297.6K[/tex]
Putting values in above equation, we get:
[tex]0.986atm\times 15.7L=n\times 0.0821\text{L atm }mol^{-1}K^{-1}\times 297.6K\\\\n=0.633mol[/tex]
To calculate the mass of bromine gas, we use the equation:
[tex]\text{Number of moles}=\frac{\text{Given mass}}{\text{Molar mass}}[/tex]
Molar mass of bromine gas = 159.8 g/mol
Moles of bromine gas = 0.633 mol
Putting values in above equation, we get:
[tex]0.633mol=\frac{\text{Mass of bromine gas}}{159.8g/mol}\\\\\text{Mass of bromine gas}=101.15g[/tex]
Hence, the mass of bromine gas contained is 101.15 grams.
