What mass of potassium hydroxide is required to react completely with 2.70 g of sulfuric acid to produce potassium sulfate and water?

Question

Wyatt Atkins · Accepted Answer

#2KOH + H_2SO_4 rarr K_2SO_4 + 2H_2O#

This problem is a bit unrealistic, as I could not really measure such a mass of sulfuric acid. Realistically, I could take a volume of sulfuric acid of a given concentration, which had an equivalent molar quantity to #(2.70*g)/(98.08*g*mol^-1)# #=# #0.0275*mol#.

Given the equation above, i.e. sulfuric acid is a diacid that requires 2 equiv of base, we require #2xx0.0275*molxx56.11*g*mol^-1=3.09*g#

If I had a #1.00*mol*L^-1# solution of sulfuric acid, what volume would I require for equivalence? Such question would be a lot more reasonable.

You undoubtedly already know that the answer to the question depends entirely on your ability to write a stoichiometric equation.

Owen Ambrose · Answer

undefined To find the mass of potassium hydroxide required, you need to determine the stoichiometry of the reaction between sulfuric acid (H2SO4) and potassium hydroxide (KOH). The balanced chemical equation for the reaction is:

H2SO4 + 2 KOH → K2SO4 + 2 H2O

From the equation, you can see that 1 mole of sulfuric acid reacts with 2 moles of potassium hydroxide. First, find the number of moles of sulfuric acid given:

Number of moles of H2SO4 = Mass of H2SO4 / Molar mass of H2SO4

Then, using the stoichiometry of the reaction, determine the number of moles of potassium hydroxide required:

Number of moles of KOH required = (Number of moles of H2SO4) * (2 moles KOH / 1 mole H2SO4)

Finally, calculate the mass of potassium hydroxide required:

Mass of KOH required = Number of moles of KOH required * Molar mass of KOH