# What mass of MnCl_2 could be isolated by reduction of MnO_2 with 150.0*g of HCl(g)?

Jan 11, 2017

Approx. $126 \cdot g$ could be isolated.

#### Explanation:

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

$\text{Hydrogen chloride}$ is the limiting reagent.

$\text{Moles of HCl}$ $=$ $\frac{150.0 \cdot g}{36.46 \cdot g \cdot m o {l}^{-} 1} = 4.11 \cdot m o l$.

Given the stoichiometry, $1.03 \cdot m o l$ $M n C {l}_{2}$ are obtained by reduction of $M n \left(I V\right)$.

This constitutes a mass of 1.03*molxx125.84*g*mol^-1=??

At one stage this reaction was used for the manufacture of chlorine. What quantity of chlorine would be produced here?