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

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Answer 1 · 2017-01-11T20:15:10

Approx. $126*g$ could be isolated.

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

$"Hydrogen chloride"$ 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?

What mass of MnCl_2MnCl_2 could be isolated by reduction of MnO_2MnO_2 with 150.0*g150.0*g of HCl(g)HCl(g)?