Mol $SO_2$=2,688/22,4=0,12 mol$M$ \rightarrow $M$+2 +2e
0,12 mol<= 0,24 mol
$S$+6 +2e \rightarrow $S$+4
0,24 mol<=0,12 mol
=>0,12M=7,68=>M=64 M là $Cu$
b) mol $O_2$=2,24/22,4=0,1 mol
Mol hhB=4,256/22,4=0,19 mol
2$SO_2$ + $O_2$ \rightleftharpoons 2$SO_3$
Bđ:0,12 mol;0,1 mol
Pứ::x mol=>0,5x mol=>x mol
Sau:0,12-x mol;0,1-0,5x mol;x mol
Molhh B=0,12-x+0,1-0,5x+x=0,19=>x=0,06 mol
Vậy hh B gồm 0,06 mol $SO2$ 0,07 mol $O2$ 0,06 mol $SO3$ =>%V
c) mol $Fe$=6,72/56=0,12 mol
Mol $Cu$=7,84/64=0,1225 mol
mcr sau pứ=8,8g>m$Cu$=>$Cu$ chưa pư vs axit, $Fe$ pứ 1phần
Fe \rightarrow $Fe$+3 +3e
x mol. => 3x mol
S+6 +2e \rightarrow S+4
3x mol=>1,5x mol
2$Fe$3+ + $Fe$ \rightarrow 3$Fe$2+
x mol=>0,5 x mol
$Fe$ dư:0,12-1,5x mol=>m$Fe$=6,72-84x
mcr=8,8=6,72-84x+7,84=>x=0,06857 mol=>mol $SO2$=0,103 mol=>V=2,304l