Question #814fe

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Answer 1 · 2016-05-23T06:01:00

This question is not well-proposed, in that sodium metal will react with the water solvent as well.

But for sodium hydroxide, #NaOH#, we can write the followig reaction:

#2NaOH(aq) + H_2SO_4(aq) rarr Na_2SO_4(aq) +2H_2O(aq)#

#"Moles of sulfuric acid"# #=# #0.750*Lxx6.0*mol*L^-1# #=# #4.50*mol#.

Thus we need to add #9.0*mol# sodium hydroxide to neutralize the sulfuric acid. #9.0*molxx40.0*g*mol^-1# #=# #360*g#.

Why do we need to add #9# #mol# caustic soda rather than #4.5# #mol#?