# Question #2b66a

May 7, 2016

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

#### Explanation:

The given equation is absolutely vital to answering this question. It says that 2 mol sodium hydroxide, approx. $80.0 \cdot g$, react with the 1 mol sulfuric acid, a mass of approx. $98.08 \cdot g$, to give stoichiometric quantities of sodium sulfate and water.

So we must calculate the molar quantities of our starting materials.

$\text{Moles of sodium hydroxide} = \frac{20.0 \cdot g}{40.00 \cdot g \cdot m o {l}^{-} 1}$ $=$ $0.5 \cdot m o l$.

$\text{Moles of sulfuric acid} = \frac{9.8 \cdot g}{98.08 \cdot g \cdot m o {l}^{-} 1}$ $=$ $0.1 \cdot m o l$.

So, clearly, sulfuric acid is the limiting reagent. And at most, I can get $0.2 \cdot m o l$ of $N a O H$ to react.

Given this, only $0.1 \cdot m o l$ of $N {a}_{2} S {O}_{4}$ will be formed, i.e. $0.1 \cdot m o l \times 142.04 \cdot g \cdot m o {l}^{-} 1$ $=$ $14.2 \cdot g$.

All of these calculations relies on the given stoichiometric reaction. If I don't have this, or if I have not balanced it correctly, I will not be able to do the calculation. Capisce?