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 $4.5$ $mol$ ?