Redox half reaction problem?

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1 Answer
Apr 2, 2018

I get around 15.2 \ "g" of lead(II) sulfate.

Explanation:

We have the following redox half-reaction, where lead (Pb) is reduced from Pb(IV) to Pb(II):

PbO_2(s)+4H^+(aq)+SO_4^(2-)(aq)+2e^(-)rightleftharpoonsPbSO_4(s)+2H_2O(l)

We are given 250 \ "mL" of 0.8 \ "M" of hydrogen ions. Let's convert that to moles.

Molarity is given by:

"molarity"="moles of solute"/"liters of solution"

And so,

0.8 \ "M"="moles of solute"/(0.25 \ "L")

"moles of solute"=0.2 \ "mol"

So, we got 0.2 moles of hydrogen ions.

According to the equation, four hydrogen ions produce one molecule of lead(II) sulfate, and so here, we would produce:

0.2color(red)cancelcolor(black)("mol of" \ H^+)*(1 \ "mol of" \ PbSO_4)/(4color(red)cancelcolor(black)("mol of" \ H^+))=0.05 \ "mol of" \ PbSO_4

Lead(II) sulfate has a molar mass of 303.26 \ "g/mol".

So here, we got:

0.05color(red)cancelcolor(black)"mol"*(303.26 \ "g")/(color(red)cancelcolor(black)"mol")=15.163 \ "g"

~~15.2 \ "g"

I rounded the answer to one decimal place, since the answer box also requires the answer to be in one decimal place.