$n_{Fe} = \dfrac{8,4}{56} = 0,15(mol)$
$3Fe + 2O_2 \xrightarrow{t^o} Fe_3O_4$
Theo PTHH : 

$n_{O_2} = \dfrac{2}{3}n_{Fe} = 0,1(mol)$
$V_{không\ khí} = \dfrac{0,1.22,4}{20\%} = 11,2(lít)$

\(3Fe+2O_2-^{t^o}\rightarrow Fe_3O_4\)

\(n_{Fe}=\dfrac{8,4}{56}=0,15\left(mol\right)\)

\(n_{O_2}=\dfrac{2}{3}n_{Fe}=0,1\left(mol\right)\)

Vì Oxi chiếm 20% thể tích không khí

=> \(V_{kk}=\dfrac{0,1.22,4}{20\%}=11,2\left(l\right)\)

a) nFe = 16,8/56 = 0,3 (mol)

PTHH: 3Fe + 2O2 -> (t°) Fe3O4

Mol: 0,3 ---> 0,2 ---> 0,1

mFe3O4 = 0,1 . 232 = 23,2 (g)

b) VO2 = 0,2 . 22,4 = 4,48 (l)

Vkk = 4,48 . 5 = 22,4 (l)

c) H = 100% - 20% = 80%

nO2 (LT) = 0,2 : 80% = 0,25 (mol)

PTHH: 2KMnO4 -> (t°) K2MnO4 + MnO2 + O2

nKMnO4 = 0,25 . 2 = 0,5 (mol)

mKMnO4 = 0,5 . 158 = 79 (g)

\(n_{CH_4}=\dfrac{3,36}{22,4}=0,15mol\)

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

0,15     0,3

\(V_{O_2}=0,3\cdot22,4=6,72l\)

\(\Rightarrow V_{kk}=5V_{O_2}=5\cdot6,72=33,6l\)

Không có đáp án đúng!

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

PTHH: 3Fe + 2O₂ --t^o--> Fe₃O₄

           0,3<---0,2-------->0,1

=> m = 0,3.56 = 16,8 (g)

b) m_Fe3O4 = 0,1.232 = 23,2 (g)

c) V_kk = 4,48 : 20% = 22,4 (l)

nO2 = 4,48/22,4 = 0,2 (mol)

PTHH: 3Fe + 2O2 -> (t°) Fe3O4

Mol: 0,3 <--- 0,2 ---> 0,1

mFe = 0,3 . 56 = 16,8 (g)

mFe3O4 = 0,1 . 232 = 23,2 (g)

Vkk = 4,48 . 5 = 22,4 (l)

a. Ag không phản ứng nên ta có PTHH: \(2Mg+O_2\rightarrow^{t^o}2MgO\)

\(\rightarrow m_{O_2}=m_{hh}-m_{\mu\text{ối}}=18,8-15,6=3,2g\)

\(\rightarrow n_{O_2}=\frac{3,2}{32}=0,1mol\)

b. \(\rightarrow V_{O_2}=n.22,4=22,4.0,1=2,24l\)

\(\rightarrow V_{kk}=4,48.5=11,2l\)

c. Có \(n_{Mg}=2n_{O_2}=0,2l\)

\(\rightarrow m_{Mg}=0,2.24=4,8g\)

\(\rightarrow\%m_{Mg}=\frac{4,8.100}{15,6}\approx30,77\%\)

\(\rightarrow\%m_{Ag}=100\%-30,77\%=69,23\%\)

\(a,PTHH:4A+3O_2\underrightarrow{t^o}2A_2O_3\\ Áp.dụng.ĐLBTKL,ta.có:\\ m_A+m_{O_2}=m_{A_2O_3}\\ \Rightarrow m_{O_2}=m_{A_2O_3}-m_A=20,4-10,8=9,6\left(g\right)\)

\(\Rightarrow n_{O_2}=\dfrac{m}{M}=\dfrac{9,6}{32}=0,3\left(mol\right)\\ Theo.PTHH:n_A=\dfrac{4}{3}.n_{O_2}=\dfrac{4}{3}.0,3=0,4\left(mol\right)\\ \Rightarrow M_A=\dfrac{m}{n}=\dfrac{10,8}{0,4}=27\left(\dfrac{g}{mol}\right)\\ \Rightarrow A.là.Al\left(nhôm\right)\)

\(b,V_{O_2\left(đktc\right)}=n.22,4=0,4.22,4=8,96\left(l\right)\\ \Rightarrow V_{kk\left(đktc\right)}=V_{O_2\left(đktc\right)}.5=8,96.5=44,8\left(l\right)\)

\(a,4A+3O_2\rightarrow\left(t^o\right)2A_2O_3\\ Theo.ĐLBTKL:\\ m_A+m_{O_2}=m_{A_2O_3}\\ \Leftrightarrow10,8+m_{O_2}=20,4\\ \Leftrightarrow m_{O_2}=9,6\left(g\right)\\ \Rightarrow n_{O_2}=\dfrac{9,6}{32}=0,3\left(mol\right)\\ n_A=\dfrac{4}{3}.0,3=0,4\left(mol\right)\Rightarrow M_A=\dfrac{m_A}{n_A}=\dfrac{10,8}{0,4}=27\left(\dfrac{g}{mol}\right)\\ \Rightarrow A:Nhôm\left(Al=27\right)\\ b,V_{kk\left(đktc\right)}=\dfrac{100}{20}.V_{O_2\left(đktc\right)}=5.\left(0,3.22,4\right)=33,6\left(l\right)\)

theo đề ta suy rar được chất sản phẩm là :\(Al_2O_3\)

a, ta có Phương trình :

\(A+O_2\underrightarrow{t^o}Al_2O_3\)

=> kim loại A là Al( nhôm)

b, \(nAl=\dfrac{10,8}{27}=0,4mol\)

pthh:

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

0,4-->0,3----->0,2

\(VO_2=0,3.24=7,2lít\)

=>\(V_{Kk}=7,2.100:20=36lít\)

PTHH: \(2K+2H_2O\rightarrow2KOH+H_2\)

\(n_K=\dfrac{7,8}{39}=0,2\left(mol\right)\)

Theo PTHH: \(n_{H_2}=\dfrac{1}{2}n_K=\dfrac{1}{2}.0,2=0,1\left(mol\right)\)

\(2H_2+O_2\xrightarrow[]{t^o}2H_2O\)

Theo PTHH: \(n_{O_2}=\dfrac{1}{2}n_{H_2}=\dfrac{1}{2}.0,1=0,05\left(mol\right)\)

\(\Rightarrow V_{O_2}=n_{O_2}.22.4=0,05.22,4=1,12\left(l\right)\)

Thể tích không khí đã dùng:

\(V_{kk}=\dfrac{V_{O_2}.100\%}{20\%}==5,6\left(l\right)\)

a, \(n_{Al}=\dfrac{5,4}{27}=0,2\left(mol\right)\)

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

Theo PT: \(n_{O_2}=\dfrac{3}{4}n_{Al}=0,15\left(mol\right)\Rightarrow V_{O_2}=0,15.24,79=3,7185\left(l\right)\)

\(\Rightarrow V_{kk}=\dfrac{V_{O_2}}{20\%}=18,5925\left(l\right)\)

b, Theo PT: \(n_{Al_2O_3}=\dfrac{1}{2}n_{Al}=0,1\left(mol\right)\Rightarrow m_{Al_2O_3}=0,1.102=10,2\left(g\right)\)

a)\(n_{Al}=\dfrac{5,4}{27}=0,2\left(m\right)\)

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

tỉ lệ       :4          3        2

số mol  :0,2       0,15   0,1

\(V_{O_2}=0,1.22,4=2,24\left(l\right)\)

\(V_{kk}=\dfrac{2,24.20\%}{100\%}=11,2\left(l\right)\)

a)\(n_{Cu}=\dfrac{3,2}{64}=0,05\left(m\right)\)

\(PTHH:2Cu+O_2\underrightarrow{t^O}2CuO\)

tỉ lệ       : 2         1       2

số mol  :0,05    0,025  0,05

\(m_{CuO}=0,05.80=4\left(g\right)\)

b)\(V_{O_2}=0,025.22,4=0,56\left(l\right)\)

\(V_{kk}=\dfrac{0,56}{20\%}=2,8\left(l\right)\)

c)\(n_{H_2}=\dfrac{2,24}{22,4}=0,1\left(m\right)\)

\(PTHH:CuO+H_2\xrightarrow[]{}Cu+H_2O\)

tỉ lệ        :1          1         1       1

số mlo   :0,1       0,1       0,1     0,1

\(m_{Cu}=0,1.64=6,4\left(g\right)\)