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

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))