a)

$4Fe + 3O_2 \xrightarrow{t^o} 2Fe_2O_3$

b)

$n_{Fe} = \dfrac{11,2}{56} = 0,2(mol) ; n_{O_2} = \dfrac{8,96}{22,4} = 0,4(mol)$

Ta thấy : 

$n_{Fe} : 4 > n_{O_2} : 3$ nên $O_2$ dư

$n_{O_2\ pư} = = \dfrac{3}{4}n_{Fe} = 0,15(mol)$
$\Rightarrow m_{O_2\ dư} = (0,4 - 0,15).32 = 8(gam)$

c) $n_{Fe_2O_3} = \dfrac{1}{2}n_{Fe} = 0,1(mol)$
$m_{Fe_2O_3} = 0,1.160 = 16(gam)$

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bn tham khảo nhé

a)

$4Al + 3O_2 \xrightarrow{t^o} 2Al_2O_3$

b) $n_{Al} = \dfrac{8,1}{27} = 0,3(mol)$

$n_{O_2} = \dfrac{13,44}{22,4} = 0,6(mol)$

Ta thấy : 

$n_{Al} : 4 < n_{O_2} : 3$ nên $O_2$ dư

$n_{O_2\ pư} = \dfrac{3}{4}n_{Al} = 0,4(mol)$
$m_{O_2\ dư} = (0,6 - 0,4).32 = 6,4(gam)$

c) $n_{Al_2O_3} = \dfrac{1}{2}n_{Al} = 0,15(mol)$
$m_{Al_2O_3} = 0,15.102 = 15,3(gam)$

Em sửa đề là : 56(g) nhé 

\(n_{Fe}=\dfrac{56}{56}=1\left(mol\right)\)

\(3Fe+2O_2\underrightarrow{t^0}Fe_3O_4\)

\(1........\dfrac{2}{3}..........\dfrac{1}{3}\)

\(m_{Fe_3O_4}=\dfrac{1}{3}\cdot232=77.33\left(g\right)\)

\(V_{O_2}=\dfrac{2}{3}\cdot22.4=14.93\left(l\right)\)

a, Ta có: \(n_{H_2}=\dfrac{6,72}{22,4}=0,3\left(mol\right)\)

\(n_{O_2}=\dfrac{4,48}{22,4}=0,2\left(mol\right)\)

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

Xét tỉ lệ: \(\dfrac{0,3}{2}< \dfrac{0,2}{1}\), ta được O₂ dư.

Theo PT: \(n_{O_2\left(pư\right)}=\dfrac{1}{2}n_{H_2}=0,15\left(mol\right)\)

\(\Rightarrow n_{O_2\left(dư\right)}=0,05\left(mol\right)\Rightarrow m_{O_2\left(dư\right)}=0,05.32=1,6\left(g\right)\)

b, \(n_{H_2O}=n_{H_2}=0,3\left(mol\right)\)

\(\Rightarrow m_{H_2O}=0,3.18=5,4\left(g\right)\)

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

_______0,3_______________________0,15 (mol)

\(\Rightarrow m_{KMnO_4}=0,3.158=47,4\left(g\right)\)

Bạn tham khảo nhé!

nKMnO4 = 9,48/158 = 0,06 (mol)

PTHH: 2KMnO4 -> (t°) K2MnO4 + MnO2 + O2 (p/ư phân hủy)

nO2 = 0,06/2 = 0,03 (mol)

nFe = 11,2/56 = 0,2 (mol)

PTHH: 3Fe + 2O2 -> (t°) Fe3O4

LTL: 0,2/3 > 0,03/4 => Fe dư

 nFe3O4 = 0,03 : 2 = 0,015 (mol)

mFe3O4 = 0,015 . 232 = 3,48 (g)

a, PT: \(3Fe+2O_2\underrightarrow{t^o}Fe_3O_4\)

b, Ta có: \(n_{Fe}=\dfrac{5,6}{56}=0,1\left(mol\right)\)

Theo PT: \(n_{Fe_3O_4}=\dfrac{1}{3}n_{Fe}=\dfrac{1}{30}\left(mol\right)\Rightarrow m_{Fe_3O_4}=\dfrac{1}{30}.232\approx7,733\left(g\right)\)

c, Theo PT: \(n_{O_2}=\dfrac{2}{3}n_{Fe}=\dfrac{1}{15}\left(mol\right)\)

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

\(n_{KMnO_4\left(LT\right)}=2n_{O_2}=\dfrac{2}{15}\left(mol\right)\)

\(\Rightarrow m_{KMnO_4\left(LT\right)}=\dfrac{2}{15}.158=\dfrac{316}{15}\left(g\right)\)

Mà: H% = 85%

\(\Rightarrow m_{KMnO_4\left(TT\right)}=\dfrac{\dfrac{316}{15}}{85\%}\approx24,78\left(g\right)\)

\(n_{H_2}=\dfrac{V_{H_2}}{22,4}=\dfrac{2,4}{22,4}\approx0,11\left(mol\right)\)

\(n_{O_2}=\dfrac{V_{O_2}}{22,4}=\dfrac{1,6}{22,4}\approx0,07\)

\(2H_2+O_2\rightarrow2H_2O\)

2 mol-1mol---2 mol

Ta có: \(\dfrac{n_{H_2}}{2}=\dfrac{0,11}{2}\)

\(\dfrac{n_{O_2}}{1}=\dfrac{0,07}{1}\)

\(\Rightarrow\dfrac{n_{H_2}}{2}< \dfrac{n_{O_2}}{1}\)

Vậy \(O_2\) dư

Số mol O₂ dư:

\(n_{O_2}=\dfrac{0,07.1}{2}=0,035\left(mol\right)\)

Khối lượng O₂ dư

\(m_{O_2}=0,035.32=1,12\left(g\right)\)

Khối lượng nước thu được:

\(n_{H_2O}=\dfrac{0,07.2}{2}=0,07\left(mol\right)\)

\(\Rightarrow m_{H_2O}=n_{H_2O}.M_{H_2O}=0,07.18=1,26\left(g\right)\)