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

Mar 24, 2017

If we assume $\text{sodium hydroxide}$ reacts with $\text{sulfuric acid}$.....approx. $90 \cdot g$ $\text{sodium sulfate}$ results......

#### Explanation:

We write the equation as:

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

$\text{Moles of sodium hydroxide} = \frac{49.9 \cdot g}{40.00 \cdot g \cdot m o {l}^{-} 1} = 1.25 \cdot m o l$.

The equation specifies that HALF an equiv of $\text{sodium sulfate}$ results from neutralization, and this represents a mass of

$1.25 \cdot m o l \times \frac{1}{2} \times 142.04 \cdot g \cdot m o {l}^{-} 1 = 88.8 \cdot g .$