nSO2 = 12,8 : 64=0,2 (mol) 
pthh : S+ O2 -t->SO2 
        0,2<--0,2<------0,2(mol) 
=> mS= 0,2.32=6,4 (g) 
=> VO2= 0,2.22,4=4,48 (l) 
ta có 
VO2 = 1/5 Vkk <=> Vkk = VO2 : 1/5 = 4,48:1/5 = 22.4 (l)

S + O₂  SO₂

S + O₂  SO₂

S+O₂-to>SO₂

0,2--0,2----0,2 mol

n _SO2=\(\dfrac{4,48}{22,4}\)=0,2 mol

=>m _S=0,2.32=6,4g

=>V_O2=0,2.22,4=4,48l

a. \(n_{SO_2}=\dfrac{4,48}{22,4}=0,2\left(mol\right)\)

PTHH : S + O₂ -t^o> SO₂

            0,2    0,2       0,2

b. \(m_S=0,2.32=6,4\left(g\right)\)

c. \(V_S=0,2.22,4=44,8\left(l\right)\)

a) PTHH :       \(S+O_2->SO_2\)

b) Ta có : \(n_S\) = \(\dfrac{m_S}{M_S}\) = 0.1 (mol)

Có :           \(n_S=n_{O_2}\)

           --> \(n_{O_2}\) = 0.1 (mol)

          => \(V_{O_2\left(đktc\right)}\) = \(n_{O_2}\) . 22.4 = 2.24 (L)

\(n_S=\dfrac{3,2}{32}=0,1\left(mol\right)\\ PTHH:S+O_2\underrightarrow{t^o}SO_2\\ \left(mol\right)..0,1\rightarrow0,1..0,1\\ V_{O_2}=0,1.22,4=2,24\left(l\right)\)

\(n_{Al}=\dfrac{5,4}{27}=0,2\left(mol\right)\\ n_{O_2}=\dfrac{50,4}{2.22,4}=0,45\left(mol\right)\)

PTHH: 4Al + 3O₂ --t^o--> 2Al₂O₃

LTL: \(\dfrac{0,2}{4}< \dfrac{0,45}{5}\rightarrow\) O₂ dư

Theo pthh: \(\left\{{}\begin{matrix}n_{Al_2O_3}=\dfrac{0,2}{2}=0,1\left(mol\right)\\n_{O_2\left(pư\right)}=\dfrac{3}{4}.0,2=0,15\left(mol\right)\end{matrix}\right.\)

\(\rightarrow\left\{{}\begin{matrix}m_{O_2\left(dư\right)}=\left(0,45-0,15\right).32=9,6\left(g\right)\\m_{Al_2O_3}=0,1.102=10,2\left(g\right)\end{matrix}\right.\)

a)`S+O_2->SO_2` nhiệt độ

`0,5-0,5`mol

`n_S=32/64=0,5`mol

`V_(O_2)=0,5.22,4=1,12l`l

c) `2KClO_3->2KCl+3O_2` nhiệt độ

`1/3-------0,5` mol

`H=80%`

`m_(KClO_3)=1/3.(122,5).80%=32,67`g

nS = 1,92/32 = 0,06 (mol)

PTHH: S + O2 -> (t°) SO2

Mol: 0,06 ---> 0,06 ---> 0,06

VSO2 (LT) = 0,06 . 22,4 = 1,344 (l)

VSO2 (TT) = 1,344 . 90% = 1,2096 (l)

nS= 0.06 mol

S + O2 -to-> SO2

Từ PTHH:

nSO2= 0.06 mol

VSO2= 1.344l

VSO2 thực = 1.344*90/100=1.2096l

nS=0,2mol

nO2=0,35mol

PTHH: S+O2=>SO2

             0,2:0,35

=> nO2 dư theo nS

p/ư:       0,2->0,2->0,2

=> V =0,2.22,4=4,48l

a, \(n_S=\dfrac{6,4}{32}=0,2\left(mol\right)\)

PTHH: S + O₂ ----t^o----> SO₂

Mol:    0,2   0,2                 0,2

b, \(m_{SO_2}=0,2.64=12,8\left(g\right)\)

c, \(V_{O_2}=0,2.22,4=4,48\left(l\right)\)

Đáp án D
S   +   O 2   →   S O 2     0 , 05             0 , 05   m o l     n S   =   1 , 6 / 32   =   0 , 05   m o l     V S O 2   =   0 , 05 . 22 , 4   =   1 , 12   l