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.
