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

May 7, 2016

#### Answer:

$2 K O H + {H}_{2} S {O}_{4} \rightarrow {K}_{2} S {O}_{4} + 2 {H}_{2} O$

#### Explanation:

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 $\frac{2.70 \cdot g}{98.08 \cdot g \cdot m o {l}^{-} 1}$ $=$ $0.0275 \cdot m o l$.

Given the equation above, i.e. sulfuric acid is a diacid that requires 2 equiv of base, we require $2 \times 0.0275 \cdot m o l \times 56.11 \cdot g \cdot m o {l}^{-} 1 = 3.09 \cdot g$

If I had a $1.00 \cdot m o l \cdot {L}^{-} 1$ solution of sulfuric acid, what volume would I require for equivalence? Such question would be a lot more reasonable.

As you no doubt realize, the ability to write a stoichiometric equation is absolutely vital to answer the question.