Question

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...?

Answer 1

Very close to `5*L`

Explanation:

`MnO_2(s) + 4HCl(aq) rarr MnCl_2(aq) + Cl_2(g)uarr + 2H_2O(l)`

`"Moles of "MnO_2` `=` `(14.75*g)/(86.94*g*mol^-1)` `=` `0.170*mol`

Given the stoichiometry of the reaction, one equiv of chlorine gas is evolved per equiv of manganese oxide consumed. Thus `0.170*mol` of `Cl_2(g)`, whose behaviour we can assume to be ideal, expresses the following volume under the given conditions of temperature and pressure.

`V=(nRT)/P=`

`((0.170*molxx0.0821*L*atm*K^-1*mol^-1xx305*K)/(646/760*atm))` `=` `?? L`

Here, I know for a fact that `1` `atm` pressure will support a column of mercury `760*mm` high. I converted the given mercury column to atmospheres in the denominator.