What mass of #"potassium sulfate"# would result from the reaction between excess #"sulfuric acid"# and #"potassium hydroxide"#?

1 Answer
anor277
May 11, 2017

Not a good question...........as it is based on a faulty premise.

Explanation:

We need a stoichiometric equation to represent the reaction:

#H_2SO_4(aq) + 2KOH(aq) rarrK_2SO_4(aq) +2H_2O(l)#

And thus moles of potassium sulfate is equivalent to half the number of moles of potassium hydroxide.

#"Moles of potassium hydroxide"# #=# #(14*g)/(56.11*g*mol^-1)#

#=0.250*mol#

And by the stoichiometry of the rxn, we get #0.125*molxx174.26*g*mol^-1=??*g# potassium sulfate, and #2xx0.125*molxx18.01*g*mol^-1=??*g# water.

The number of molecules is simply the molar quantity multiplied by #N_A, "Avogadro's number", 6.022xx10^23*mol^-1#.

But in fact, all of these calculations are based on a FAULTY premise. The product of potassium hydroxide with EXCESS sulfuric acid is not #"potassium sulfate"#, #K_2SO_4#, but #"potassium bisulfate"#, i.e. #KHSO_4(aq)#. In the reaction as written we would get 1 equiv of #KHSO_4(aq)# and ONE of water:

#H_2SO_4(aq) + KOH(aq) rarr KHSO_4(aq) + H_2O(l)#.

So if you want to be bolshie, you can state that NO potassium sulfate results from the reaction.

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