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

\(n_{O_2}=\dfrac{11,2}{22,4}=0,5\left(mol\right)\)

PTHH: S + O₂ --t^o--> SO₂

Xét tỉ lệ: \(\dfrac{0,1}{1}< \dfrac{0,5}{1}\) => S hết, O₂ dư

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

\(n_{O_2}=\dfrac{11,2}{22,4}=0,5\left(mol\right)\)

PTHH: S + O₂ --t^o--> SO₂

Xét tỉ lệ: \(\dfrac{0,1}{1}< \dfrac{0,5}{1}\) => S hết, O₂ dư

.

\(n_S=\dfrac{3,2}{32}=0,1\left(mol\right)\\ n_{O_2}=\dfrac{1,12}{22,4}=0,05\left(mol\right)\\ pthh:S+O_2\underrightarrow{t^o}SO_2\\ LTL:\dfrac{0,1}{1}>\dfrac{0,05}{1}\) 
=> S dư 
\(n_{S\left(P\text{Ư}\right)}=n_{SO_2}=n_{O_2}=0,05\left(mol\right)\\ m_S=\left(0,1-0,05\right).32=1,6\left(g\right)\\ V_{SO_2}=0,05.22,4=1,12\left(l\right)\)

\(n_{SO_2}=\dfrac{2.24}{22.4}=0.1\left(mol\right)\)

\(S+O_2\underrightarrow{^{t^0}}SO_2\)

\(n_S=0.1\left(mol\right)\)

\(m_S=0.1\cdot32=3.2\left(g\right)\)

=> A 

PTHH : S + O2 -> SO2

nSO2 = V/22,4= 0,1 mol

Theo PTHH : nS = nSO2 = 0,1 mol

=> mS = n.M = 3,2 g

nS = 6,4/32 = 0,2 (mol)

nO2 = 2,24/22,4 = 0,1 (mol)

PTHH: S + O2 -> (t°) SO2

LTL: 0,2 > 0,1 => S dư

nS (p/ư) = nSO2 = nO2 = 0,1 (mol)

=> VSO2 = 0,1 . 22,4 = 2,24 (l)

=> mS (dư) = (0,2 - 0,1) . 32 = 3,2 (g)

cảm ơn bạn cảm ơn bạn rất rất rất nhiều luôn <:3

20 % j đó k có dùng á ???

a, PTHH: S + O2 -> (t°) SO2

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

nO2 = 6,72/22,4 = 0,3 (mol)

LTL: 0,2 < 0,3 => O2 dư

nO2 (pư) = nSO2 = nS = 0,2 (mol)

mO2 (dư) = (0,3 - 0,2) . 32 = 3,2 (g)

c, mSO2 = 64 . 0,2 = 12,8 (g)

a, \(S+O_2\underrightarrow{t^o}SO_2\)

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

\(nO_2=\dfrac{6,72}{22,4}=0,3\left(mol\right)\)

\(\dfrac{0,2}{1}< \dfrac{0,3}{1}\)  => oxi dư 

\(nO_{2\left(dư\right)}=0,1\left(mol\right)\)

\(mO_{2\left(dư\right)}=0,1.32=3,2\left(g\right)\)

\(nSO_2=nS=0,2\left(mol\right)\)

\(mSO_2=0,2.64=12,8\left(g\right)\)

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

\(n_S=\dfrac{3.2}{32}=0.1\left(mol\right)\)

\(n_{O_2}=\dfrac{2.688}{22.4}=0.12\left(mol\right)\)

\(S+O_2\underrightarrow{t^0}SO_2\)

\(0.1...0.1.....01\)

\(V_{O_2\left(dư\right)}=\left(0.12-0.1\right)\cdot22.4=0.448\left(l\right)\)

\(V_{SO_2}=0.1\cdot22.4=2.24\left(l\right)\)

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

\(n_{O_2}=\dfrac{2,688}{22,4}=0,12\left(mol\right)\)

PTHH :                 \(S+O_2\rightarrow SO_2\)

Ban đầu :           0,1    0,12              (mol)

Phản ứng :        0,1    0,1      0,1     (mol)

Sau phản ứng :  0       0,02    0,1     (mol)

\(m_{O_2}=0,02.32=0,64\left(g\right)\)

\(m_{SO_2}=0,1.64=6,4\left(g\right)\)

\(V_{SO_2}=0,1.22,4=2,24\left(l\right)\)