# How many milliliters of 2.00 M H_2SO_4 will react with 28.0 g of NaOH?

Jun 11, 2016

$175 \cdot m L$ of the given sulfuric acid.

#### Explanation:

We need (i) a balanced chemical equation:

${H}_{2} S {O}_{4} \left(a q\right) + 2 N a O H \left(a q\right) \rightarrow N {a}_{2} S {O}_{4} \left(a q\right) + 2 {H}_{2} O \left(l\right)$

And (ii) equivalent molar quantitites:

$\text{Moles of NaOH}$ $=$ $\frac{28.0 \cdot g}{40.0 \cdot g \cdot m o {l}^{-} 1}$ $=$ $0.70 \cdot m o l$

Given the stoichiometry of the reaction, we require a half equivalent of sulfuric acid, i.e. $0.35 \cdot m o l$.

So we divide this molar quantity by the concentration to get an answer in $m L$.

$\text{Volume}$ $=$ $\frac{0.35 \cdot m o l}{2.0 \cdot m o l \cdot {L}^{-} 1} \times 1000 \cdot m L \cdot {L}^{-} 1$ $=$ $175 \cdot m L$.