Sodium reacts with chlorine gas according to the following reaction: 2Na (s) + Cl2 (g) →2NaCl (s) What volume of Cl2 gas, measured at 689 torr and 39 ∘C, is required to form 28 g of NaCl?

Answer:6.719Litres of Cl2 gas.

Explanation:According to eqn of rxn

2Na + Cl2=2NaCl

P=689torr=689/760=0.91atm

T=39°C+273=312K

according to stoichiometry of the reaction, 1Moles of Cl2 gives 2moles of NaCl

But 28g of NaCl was given, we have to convert this to moles by using the relation, n=mass/MW

MW of NaCl=23+35.5=58.5g/mol

n=28g (mass given of NaCl) / 58.5

n=0.479moles of NaCl

Going back to the reaction,

if 1moles of Cl2 produces 2moles of NaCl

x moles of Cl2 will give 0.479moles of NaCl.

x=0.479*1/2

x=0.239moles of Cl2.

To find the volume, we use ideal ggas eqn, PV=nRT

V=nRT/P

V=0.239*0.082*312/0.91

V=6.719Litres