What volume of #Cl_2# gas, measured at 684 torr and 32 C, is required to form 26 g of #NaCl#?

1 Answer
Dec 3, 2016

Answer:

We require a volume of #~=6*L# #Cl_2# gas.

Explanation:

#Na(s)+1/2Cl_2(g) rarrNaCl(s)#

#"Moles of NaCl"# #=# #(26.0*g)/(58.44*g*mol^-1)=0.445*mol#.

And thus we need half an equiv of #Cl_2# gas, i.e. #0.222*mol#.

If we use the Ideal Gas law, then #V=(nRT)/P#, and also we know that #760*mm*Hg-=1*atm#.

And thus #V=(0.222*molxx0.0821*L*atm*K^-1*mol^-1xx305*K)/((684*mm*Hg)/(760*mm*Hg*atm^-1))#