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

1 Answer
anor277
Jan 11, 2017

Approx. #126*g# could be isolated.

Explanation:

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

#"Hydrogen chloride"# is the limiting reagent.

#"Moles of HCl"# #=# #(150.0*g)/(36.46*g*mol^-1)=4.11*mol#.

Given the stoichiometry, #1.03*mol# #MnCl_2# are obtained by reduction of #Mn(IV)#.

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?

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