# Upon treatment of a 14.75*g mass of MnO_2 with excess HCl(aq), what VOLUME of chlorine gas is generated under standard conditions...?

May 2, 2016

Very close to $5 \cdot L$

#### Explanation:

$M n {O}_{2} \left(s\right) + 4 H C l \left(a q\right) \rightarrow M n C {l}_{2} \left(a q\right) + C {l}_{2} \left(g\right) \uparrow + 2 {H}_{2} O \left(l\right)$

$\text{Moles of } M n {O}_{2}$ $=$ $\frac{14.75 \cdot g}{86.94 \cdot g \cdot m o {l}^{-} 1}$ $=$ $0.170 \cdot m o l$

Given the stoichiometry of the reaction, one equiv of chlorine gas is evolved per equiv of manganese oxide consumed. Thus $0.170 \cdot m o l$ of $C {l}_{2} \left(g\right)$, whose behaviour we can assume to be ideal, expresses the following volume under the given conditions of temperature and pressure.

$V = \frac{n R T}{P} =$

$\left(\frac{0.170 \cdot m o l \times 0.0821 \cdot L \cdot a t m \cdot {K}^{-} 1 \cdot m o {l}^{-} 1 \times 305 \cdot K}{\frac{646}{760} \cdot a t m}\right)$ $=$ ?? L

Here, I know for a fact that $1$ $a t m$ pressure will support a column of mercury $760 \cdot m m$ high. I converted the given mercury column to atmospheres in the denominator.