# What volume of Cl_2 gas, measured at 684 torr and 32 C, is required to form 26 g of NaCl?

Dec 3, 2016

We require a volume of $\cong 6 \cdot L$ $C {l}_{2}$ gas.

#### Explanation:

$N a \left(s\right) + \frac{1}{2} C {l}_{2} \left(g\right) \rightarrow N a C l \left(s\right)$

$\text{Moles of NaCl}$ $=$ $\frac{26.0 \cdot g}{58.44 \cdot g \cdot m o {l}^{-} 1} = 0.445 \cdot m o l$.

And thus we need half an equiv of $C {l}_{2}$ gas, i.e. $0.222 \cdot m o l$.

If we use the Ideal Gas law, then $V = \frac{n R T}{P}$, and also we know that $760 \cdot m m \cdot H g \equiv 1 \cdot a t m$.

And thus $V = \frac{0.222 \cdot m o l \times 0.0821 \cdot L \cdot a t m \cdot {K}^{-} 1 \cdot m o {l}^{-} 1 \times 305 \cdot K}{\frac{684 \cdot m m \cdot H g}{760 \cdot m m \cdot H g \cdot a t {m}^{-} 1}}$