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...?

1 Answer
May 2, 2016

Answer:

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.