Question

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

Answer 1

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))`