The balanced equation is
#"2Fe + 3Cl"_2 → "2FeCl"_3#
1. Calculate the moles of Fe
#"Moles of Fe" = 7000 color(red)(cancel(color(black)("g Fe"))) × "1 mol Fe"/(55.84 color(red)(cancel(color(black)("g Fe")))) = "125.4 mol Fe"#
2. Calculate moles of #"Cl"_2#
#"Moles of Cl"_2 = 125.4 color(red)(cancel(color(black)("mol Fe"))) × ("3 mol Cl"_2)/(2 color(red)(cancel(color(black)("mol Fe")))) = "188.0 mol Cl"_2"#
3. Calculate the volume of #"Cl"_2#
We can use the Ideal Gas Law:
#color(blue)(|bar(ul(color(white)(a/a)PV = nRT color(white)(a/a)|)))" "#
Rearrangement gives us
#V = (nRT)/P#
Your data are:
n = #"188.0 mol"#; #color(white)(mmmmmmmmmll)R "= 8.314 kPa·L·K"^"-1""mol"^"-1"#;
#T = "(20 + 273.15) K = 293.15 K"#; #P = "120 kPa"#
∴ #V = (188.0 color(red)(cancel(color(black)("mol"))) × 8.314 color(red)(cancel(color(black)("kPa")))·"L"·color(red)(cancel(color(black)("K"^"-1""mol"^"-1"))) × 293.15 color(red)(cancel(color(black)("K"))))/(120 color(red)(cancel(color(black)("kPa")))) = "3720 L" = "372 m"^3#
The volume of chlorine is #"372 m"^3#.
