a) 

- \(V_{CO}=n.24=0,2.24=4,8\left(l\right)\)

- \(n_{SO_3}=\dfrac{m}{M}=\dfrac{8}{80}=0,1\left(mol\right)\)

`=>` \(V_{SO_3}=n.24=0,1.24=2,4\left(l\right)\)

- \(n_{N_2}=\dfrac{\text{Số phân tử}}{6.10^{23}}=\dfrac{3.10^{23}}{6.10^{23}}=0,5\left(mol\right)\)

`=>` \(V_{N_2}=n.24=0,5.24=12\left(l\right)\)

b) 

- \(m_{Fe_2O_3}=n.M=0,25.160=40\left(g\right)\)

- \(m_{Al_2O_3}=n.M=0,15.102=15,3\left(g\right)\)

- \(n_{O_2}=\dfrac{V_{\left(\text{đ}ktc\right)}}{22,4}=\dfrac{3,36}{22,4}=0,15\left(mol\right)\)

`=>` \(m_{O_2}=n.M=0,15.32=4,8\left(g\right)\)

c) 

Ta có: \(\left\{{}\begin{matrix}n_{SO_2}=\dfrac{m}{M}=\dfrac{8}{64}=0,125\left(mol\right)\\n_{CO_2}=\dfrac{m}{M}=\dfrac{4,4}{44}=0,1\left(mol\right)\\n_{H_2}=\dfrac{m}{M}=\dfrac{0,1}{2}=0,05\left(mol\right)\end{matrix}\right.\)

`=>` \(n_{hh}=n_{SO_2}+n_{CO_2}+n_{H_2}=0,125+0,1+0,05=0,275\left(mol\right)\)

`=>` \(V_{hh\left(\text{đ}ktc\right)}=n_{hh}.22,4=0,275.22,4=6,16\left(l\right)\)

a. \(V_{NO_2}=0,357.22,4=7,9968\left(lít\right)\)

b. \(V_{SO_3}=\dfrac{4}{80}.22,4=1,12\left(lít\right)\)

c. \(V_{H_2}=\dfrac{5,4.10^{24}}{6.10^{23}}.22,4=201,6\left(lít\right)\)

\(n_{SO3}=\dfrac{6,022.10^{23}}{6,022.10^{23}}=1\left(mol\right)\)

a) \(V_{SO3}=1.24,79=24,79\left(l\right)\)

b) \(d_{SO3/O2}=\dfrac{80}{32}=2,5\) ⇒ SO₃ nặng gấp 2,5 lần O₂

    \(d_{SO3/kk}=\dfrac{80}{29}\approx2,76\) ⇒ SO₃ nặng gấp 2,76 lần không khí

 Chúc bạn học tốt  

\(n_{SO_3}=\dfrac{m}{M}=\dfrac{4}{80}=0,05\left(mol\right)\)

⇒ \(V_{SO_3\left(đktc\right)}=n.22,4=0,05.22,4=1,12\left(l\right)\)

\(n_{CH_4}=\dfrac{9.10^{23}}{6.10^{23}}=1,5\left(mol\right)\)

⇒ \(V_{CH_4\left(đktc\right)}=n.22,4=1,5.22,4=33,6\left(l\right)\)

\(2SO_2+O_2\xrightarrow[V_2O_5]{t^o}2SO_3\\ n_{SO_2}=\dfrac{6,4}{64}=0,1(mol)\\ \Rightarrow n_{O_2}=0,05(mol);n_{SO_3}=0,1(mol)\\ \Rightarrow V_{O_2}=0,05.22,4=1,12(lít);V_{SO_3}=0,1.22,4=2,24(lít)\)

a) \(V_{SO_3}=0,2.22,4=4,48\left(l\right)\)

b) \(m_{SO_3}=0,2.80=16\left(g\right)\)

c) Số phân tử SO₃ = \(0,2.6.10^{23}=1,2.10^{23}\)