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. \(V_S=0,2.22,4=44,8\left(l\right)\)

n_S = 9,6/32 = 0,3 mol

S + O₂ ---t^o----> SO₂

0,3__0,3__________0,3

m_SO2 = 0,3 . 64 = 19,2 (g)

V_O2 = 0,3 . 22,4 = 6,72 (l)

V_kk = 6,72 . 5 = 33,6 (l)

\(n_S=\dfrac{9,6}{32}=0,3\left(mol\right)\\ pthh:S+O_2\underrightarrow{t^o}SO_2\) 
         0,3   0,3    0,3 
\(m_{SO_2}=0,3.64=19,2\left(g\right)\\ V_{KK}=\left(0,3.22,4\right):\dfrac{1}{5}=33,6\left(L\right)\)

\(n_S=\dfrac{9,6}{32}=0,3\left(mol\right)\)

PTHH: S + O₂ --t^o--> SO₂

          0,3-->0,3----->0,3

\(\rightarrow\left\{{}\begin{matrix}m_{SO_2}=0,3.64=19,2\left(g\right)\\V_{kk}=0,3.5.22,4=33,6\left(l\right)\end{matrix}\right.\)

n_S = \(\dfrac{3,2}{32}=0,1mol\)

a)

PTHH: S + O₂ -^to--> SO₂

             _{0,1     0,1        0,1   (mol)}

b) m_SO2 = 0,1.64 = 6,4g

c) V_O2 = 0,1.22,4 = 2,24 lít

Vkk = 5.V_O2 = 5.2,24 = 11,2 lít

a)\(n_S=\dfrac{3,2}{32}=0,1(mol)\)

\(PTHH:S+O_2\xrightarrow{t^o}SO_2\)

\(PT:1-1-1\)

\(Đề:0,1-/-/\)

\(PƯ:0,1\rightarrow0,1\rightarrow0,1\)

b) \(n_{SO_2}=0,1\Rightarrow m_{SO_2}=0,1.64=6,4(g)\)

c)

\(n_{O_2}=0,1\Rightarrow V_{O_2}=22,4.0,1=2,24(l)\)

\(V_{kk}=5.V_{O_2}=5.2,24=11,2(l)\)

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)

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

a, \(n_S=\dfrac{12,8}{32}=0,4\left(mol\right)\\ n_{O_2}=\dfrac{11,2}{22,4}=0,5\left(mol\right)\)

PTHH: S + O₂ --t^o--> SO₂

LTL: 0,4 < 0,5 => khí oxi dư

b, \(\left\{{}\begin{matrix}n_{SO_2}=0,4\left(mol\right)\\n_{O_2\left(pư\right)}=0,4\left(mol\right)\end{matrix}\right.\\ \Rightarrow V=\left(0,4+0,5-0,4\right).22,4=11,2\left(l\right)\)

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

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

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

a. PTHH: S + O₂ ---t^o---> SO₂

Theo PT: \(n_{SO_2}=n_S=0,2\left(mol\right)\)

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

b. Theo PT: \(n_{O_2}=n_S=0,2\left(mol\right)\)

=> \(m_{O_2}=0,2.32=6,4\left(g\right)\)

a)S+O2-------->SO2

b)n S=6,4/32=0,2(mol)

Theo pthh

n SO2 =n S=0,2(mol)

V SO2=0,2.22,4=4,48(mol)