n_S=m_S/M_S=3,2/32=0,1(mol)
      n_O2=V_O2/22,4=32/22,4=1,42(mol)
PTHH: S + O₂ --> SO₂ (1)
BĐ:    0,1   1,42
PỨ:   0,1-->0,1-->0,1
SPỨ:  0--->0,32-->0,1
a) Từ PT(1)=>O₂ dư
_VO2(dư)=n_O2(dư) .22,4=0,32 .22,4=7,168(l)
b) Từ PT(1)=>n_SO2=0,1(mol)
=>m_SO2=n.M=0,1 .64=6,4(g)

PTHH; CH4 + 2O2 → 2H2O + CO2↑

nCH4=3,2\16=0,2(mol)

Theo PTHH, ta có: nO2=2nCH4=2.0,2=0,4(mol))

⇒VO2=0,4.22,4=8,96(l)

Theo PTHH, ta có:nCO2=nCH4=0,2(mol)

⇒mCO2=0,2.48=9,6(g)

\(n_{CH_4}=\dfrac{3.2}{16}=0.2\left(mol\right)\)

\(CH_4+2O_2\underrightarrow{t^0}CO_2+2H_2O\)

\(0.2........0.4........0.2\)

\(V_{O_2}=0.4\cdot22.4=8.96\left(l\right)\)

\(m_{CO_2}=0.2\cdot44=8.8\left(g\right)\)

a) \(n_{O_2}=\dfrac{22,4.20\%}{22,4}=0,2\left(mol\right)\)

Gọi số mol Al, Mg là a, b (mol)

=> 27a + 24b = 7,8 (1)

PTHH: 4Al + 3O₂ --t^o--> 2Al₂O₃

             a-->0,75a------>0,5a

             2Mg + O₂ --t^o--> 2MgO

             b-->0,5b------->b

=> 0,75a + 0,5b = 0,2 (2)

(1)(2) => a = 0,2; b = 0,1

=> \(\left\{{}\begin{matrix}m_{Al}=0,2.27=5,4\left(g\right)\\m_{Mg}=0,1.24=2,4\left(g\right)\end{matrix}\right.\)

=> \(\left\{{}\begin{matrix}\%m_{Al}=\dfrac{5,4}{7,8}.100\%=69,23\%\\\%m_{Mg}=\dfrac{2,4}{7,8}.100\%=30,77\%\end{matrix}\right.\)

b)

C1: m_{hh oxit} = 7,8 + 0,2.32 = 14,2 (g)

C2: \(\left\{{}\begin{matrix}m_{Al_2O_3}=0,1.102=10,2\left(g\right)\\m_{MgO}=0,1.40=4\left(g\right)\end{matrix}\right.\)

=> m_{hh oxit} = 10,2 + 4 = 14,2 (g)

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)