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

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Answer 1 · 2017-03-24T17:13:14

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

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

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

If a 49.9*g49.9*g mass of sodium hydroxide reacts with sulfuric acid, what mass of sodium sulfate results?