If a #49.9*g# mass of sodium hydroxide reacts with sulfuric acid, what mass of sodium sulfate results?

1 Answer
Mar 24, 2017

Answer:

If we assume #"sodium hydroxide"# reacts with #"sulfuric acid"#.....approx. #90*g# #"sodium sulfate"# results......

Explanation:

We write the equation as:

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

#"Moles of sodium hydroxide"=(49.9*g)/(40.00*g*mol^-1)=1.25*mol#.

The equation specifies that HALF an equiv of #"sodium sulfate"# results from neutralization, and this represents a mass of

#1.25*molxx1/2xx142.04*g*mol^-1=88.8*g.#