a, PT: \(4Al+3O_2\underrightarrow{t^o}2Al_2O_3\)

Ta có: \(n_{Al}=\dfrac{10,2}{27}=\dfrac{17}{45}\left(mol\right)\)

b, Theo PT: \(n_{O_2}=\dfrac{3}{4}n_{Al}=\dfrac{17}{60}\left(mol\right)\)

\(\Rightarrow V_{O_2}=\dfrac{17}{60}.22,4\approx6,347\left(l\right)\)

c, Theo PT: \(n_{Al_2O_3}=\dfrac{1}{2}n_{Al}=\dfrac{17}{90}\left(mol\right)\)

\(\Rightarrow m_{Al_2O_3}=\dfrac{17}{90}.102\approx19,267\left(g\right)\)

d, PT: \(2KMnO_4\underrightarrow{t^o}K_2MnO_4+MnO_2+O_2\)

Theo PT: \(n_{KMnO_4}=2n_{O_2}=\dfrac{17}{30}\left(mol\right)\)

\(\Rightarrow m_{KMnO_3}=\dfrac{17}{30}.158\approx89,53\left(g\right)\)

Theo PTHH

\(n_{H2}=n_{H20}=\dfrac{3,6}{18}=0,2(mol)\)

\(V_{H2}=0,2 . 22,4 =4,48 (lít)\)

Theo PTHH :

⇒\(V_{H2}\)= 0,2 . 22,4 = 4,48(lít)

a) C + O₂ --to--> CO₂

b) \(n_C=\dfrac{3}{12}=0,25\left(mol\right)\)

PTHH: C + O₂ --to--> CO₂
___0,25->0,25----->0,25________(mol)

=> V_O2 = 0,25.22,4 = 5,6(l)

c) VCO2  = 0,25.22,4 = 5,6 (l)

\(n_{CO_2}=\dfrac{4}{12}=0,25(mol)\\ a,PTHH:C+O_2\xrightarrow{t^o}CO_2\\ b,n_{O_2}=n_{C}=0,25(mol)\\ \Rightarrow V_{O_2}=0,25.22,4=5,6(l)\\ \text{Do }O_2\text{ chiếm }\dfrac{1}{5}\text{ thể tích kk}\\ \Rightarrow V_{kk}=5,6.5=28(l)\)

\(c,n_{CO_2}=n_{C}=0,25(mol)\\ \Rightarrow V_{CO_2}=0,25.22,4=5,6(l)\)

Ta có: \(n_{Fe_2O_3}=\dfrac{48}{160}=0,3\left(mol\right)\)

PT: \(Fe_2O_3+3H_2\underrightarrow{t^o}2Fe+3H_2O\)

_____0,3_____0,9___0,6____0,9 (mol)

a, \(m_{Fe}=0,6.56=33,6\left(g\right)\)

b, \(V_{H_2}=0,9.22,4=20,16\left(l\right)\)

c, PT: \(2H_2+O_2\underrightarrow{t^o}2H_2O\)

Theo PT: \(n_{O_2}=\dfrac{1}{2}n_{H_2O}=0,9\left(mol\right)\)

\(\Rightarrow V_{O_2}=0,9.22,4=20,16\left(l\right)\)

a, Theo giả thiết ta có: \(n_P=\dfrac{3,1}{31}=0,1\left(mol\right)\)

\(4P+5O_2--t^o->2P_2O_5\)

Ta có: \(n_{O_2}=\dfrac{5}{4}.n_P=0,125\left(mol\right)\Rightarrow V_{O_2\left(đktc\right)}=0,125.22,4=2,8\left(l\right)\)

b, Theo giả thiết ta có: \(n_{CH_4}=\dfrac{1,12}{22,4}=0,05\left(mol\right)\)

\(CH_4+2O_2--t^o->CO_2+2H_2O\)

Ta có: \(n_{O_2}=2.n_{CH_4}=0,1\left(mol\right)\Rightarrow V_{O_2\left(đktc\right)}=2,24\left(l\right)\)

a)\(n_{H_2}=\dfrac{8,96}{22,4}=0,4mol\)

\(2H_2+O_2\underrightarrow{t^o}2H_2O\)

 0,4      0,2     0,4

\(V_{O_2}=0,2\cdot22,4=4,48l\)

\(V_{kk}=5V_{O_2}=5\cdot4,48=22,4l\)

b)\(CuO+H_2\rightarrow Cu+H_2O\)

                0,4                0,4

\(m_{H_2O}=0,4\cdot18=7,2g\)

nH2=6,72/22,4=0,3 mol

2H2 +O2 -to-> 2H2O 

0,3     0,15              mol

VO2 =0,15*22,4=3,36 lít

V không Khí = 3,3685=16,8 g

nH2O=0,3  mol

mH2O=0,3*18=5,4 g

Fe + H2So4 --> FeSO4 + H2

0,3                                     0,3    mol

=>mFe=0,3 *56=16,8 g