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Answer 1 · 2015-05-30T00:26:07

The volume of chlorine gas is #"60.2 m"^3#.

We start with the balanced equation.

#"2NaCl" → "2Na" + "Cl"_2#

Step 1. Calculate the moles of #"Na"#.

#105 cancel("kg Na") × "1 kmol Na"/(22.99 cancel("kg Na")) = "4.567 kmol Na"#

Step 2. Calculate the moles of #"Cl"_2#.

The balanced equation tells us that 1 kmol of #"Cl"_2# is formed for every 2 kmol of #"NaCl"#. So,

#4.567 cancel("kmol Na") × ("1 kmol Cl"_2)/(2 cancel("kmol Na")) = "2.284 kmol Cl"_2#

Step3. To calculate the volume, we use the Ideal Gas Law:

#PV = nRT#

#V = (nRT)/P = (2284 cancel("mol") × "8.314L"·cancel("kPa·K⁻¹mol⁻¹") × 303.15 cancel("K"))/(95.7 cancel("kPa")) = "60 100 L" = "60.1 m"^3#