Ta có: \(n_{C_2H_4}=\dfrac{11,2}{22,4}=0,5\left(mol\right)\)

PT: \(C_2H_4+3O_2\underrightarrow{t^o}2CO_2+2H_2O\)

a, \(n_{O_2}=3n_{C_2H_4}=1,5\left(mol\right)\Rightarrow V_{O_2}=1,5.22,4=33,6\left(l\right)\)

\(\Rightarrow V_{kk}=\dfrac{V_{O_2}}{20\%}=168\left(l\right)\)

b, \(n_{CO_2}=2n_{C_2H_4}=1\left(mol\right)\Rightarrow m_{CO_2}=1.44=44\left(g\right)\)

a, PT: \(2Al+6HCl\rightarrow2AlCl_3+3H_2\)

Ta có: \(n_{Al}=\dfrac{2,7}{27}=0,1\left(mol\right)\)

Xét tỉ lệ: \(\dfrac{0,1}{2}>\dfrac{0,25}{6}\), ta được Al dư.

Theo PT: \(n_{H_2}=\dfrac{1}{2}n_{HCl}=0,125\left(mol\right)\Rightarrow V_{H_2}=0,125.22,4=2,8\left(l\right)\)

b, Theo PT: \(n_{Al\left(pư\right)}=\dfrac{1}{3}n_{HCl}=\dfrac{1}{12}\left(mol\right)\Rightarrow n_{Al\left(dư\right)}=0,1-\dfrac{1}{12}=\dfrac{1}{60}\left(mol\right)\)

\(\Rightarrow m_{Al}=\dfrac{1}{60}.27=0,45\left(g\right)\)

a) 

\(n_{Al}=\dfrac{13,5}{27}=0,5\left(mol\right)\)

PTHH: 4Al + 3O₂ --t^o-- 2Al₂O₃

_____0,5-->0,375--->0,25

=> m_Al2O3 = 0,25.102 = 25,5 (g)

b) V_O2 = 0,375.22,4 = 8,4 (l)

=> V_kk = 8,4.5 = 42 (l)

4Al+3O2-to>2Al2O3

0,2----0,15-------0,1

nAl=\(\dfrac{5,4}{27}\)=0,2 mol

m Al2O3=0,1.102=10,2g

=>VO2=0,15.22,4=3,36l

2KClO3-to>2KCl+3O2

0,1----------------------0,15

=>m KClO3=0,1.122,5=12,25g

\(n_{Al}=\dfrac{m}{M}=\dfrac{13,5}{27}=0,5\left(mol\right)\)

PTHH:\(4Al+3O_2\rightarrow2Al_2O_3\)

             0,5    0,375     0,25      (mol)

a)\(V_{O_2}=n.22,4\)=0,375.22,4=8,4(l)

b)\(m_{Al_2O_3}=n.M\)=0,25.102=25,5(g)

\(n_{Al}=\dfrac{13,5}{27}=0,5\left(mol\right)\)

Pt : \(4Al+3O_2\underrightarrow{t^o}2Al_2O_3|\)

        4         3            2

      0,5                    0,25

Câu a hình đề bị sai bạn xem lại giúp mình 

b) \(n_{Al2O3}=\dfrac{0,5.2}{4}=0,25\left(mol\right)\)

⇒ \(m_{Al2O3}=0,25.102=25,5\left(g\right)\)

 Chúc bạn học tốt

*Sửa đề: Tính thể tích Oxi cần dùng

PTHH: \(4Al+3O_2\xrightarrow[]{t^o}2Al_2O_3\)

Ta có: \(n_{Al}=\dfrac{13,5}{27}=0,5\left(mol\right)\)

\(\Rightarrow\left\{{}\begin{matrix}n_{O_2}=0,375\left(mol\right)\\n_{Al_2O_3}=0,25\left(mol\right)\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}V_{O_2}=0,375\cdot22,4=8,4\left(l\right)\\m_{Al_2O_3}=0,25\cdot102=25,5\left(g\right)\end{matrix}\right.\)

\(n_{Al_2O_3}=\dfrac{20,4}{102}=0,2\left(mol\right)\)

\(a)PT:4Al+3O_2\xrightarrow[]{t^o}2Al_2O_3\)

\(b)tỉ.lệ:4...........3...........2\)

theo PT: \(4mol..........3mol...........2mol\)

                 \(x..........y..........0,2\)

\(\rightarrow n_{O_2}=\dfrac{0,2.3}{2}=0,3\left(mol\right)\)

\(\rightarrow n_{Al}=\)\(\dfrac{0,2.4}{2}=0,4\left(mol\right)\)

\(c)V_{O_2\left(đktc\right)}=n_{O_2}.22,4=0,3.22,4=6,72\left(l\right)\)

\(d)d_{O_2/CH_4}=\dfrac{M_{O_2}}{M_{CH_4}}=\dfrac{32}{16}=2.lần\)

Đáp án A

Hỗn hợp rắn gồm Al, Al₂O₃, Fe với n Fe = 2 n Al 2 O 3 . Chia thành 2 phần không bằng nhau: