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

Answer 1

Wyatt Atkins

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?

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Answer 2

Cameron Andrews

To calculate the mass of MnCl2 isolated by the reduction of MnO2 with HCl, you need to first balance the chemical equation for the reaction:

MnO2 + 4 HCl -> MnCl2 + 2 H2O + Cl2

Then, use stoichiometry to determine the mass of MnCl2 produced. First, calculate the moles of HCl using its molar mass and mass given. Then, using the stoichiometric ratio from the balanced equation, determine the moles of MnCl2 produced. Finally, convert moles of MnCl2 to grams using its molar mass.

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Answer from HIX Tutor

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.