What mass of salt can be isolated from a #75*g# mass of sodium metal, and a #25*g# mass of chlorine gas?

1 Answer
Mar 22, 2017

Answer:

Approx. #41*g# of salt will be produced.

Explanation:

We interrogate the reaction:

#Na(s) + 1/2Cl_2(g) rarr NaCl(s)#

One equiv of sodium metal reacts with half an equiv of #Cl_2(g)# to give one equiv of salt.

#"Moles of sodium"=(75*g)/(22.99*g*mol^-1)=3.26*mol#

#"Moles of chlorine gas"=(25*g)/(70.90*g*mol^-1)=0.353*mol#.

So sodium metal is present in wast stoichiometric excess, and chlorine gas is the limiting reagent.

And thus AT MOST, only #2xx0.353*mol# #NaCl# can be generated, i.e. #2xx0.353*cancel(mol)xx58.44*g*cancel(mol^-1)=41.3*g#