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)

$a\big)$

$n_{Fe}=\frac{16,8}{56}=0,3(mol)$

$3Fe+2O_2\xrightarrow{t^o}Fe_3O_4$

Theo PT: $n_{Fe_3O_4}=\frac{1}{3}n_{Fe}=0,1(mol)$

$\to m_{Fe_3O_4}=0,1.232=23,2(g)$

$b\big)$

Theo PT: $n_{O_2}=\frac{2}{3}n_{Fe}=0,2(mol)$

$\to V_{O_2}=0,2.22,4=4,48(l)$

$\to V_{kk}=4,48.5=22,4(l)$

$c\big)$

$2KMnO_4\xrightarrow{t^o}K_2MnO_4+MnO_2+O_2$

Theo PT: $n_{KMnO_4}=2n_{O_2}=0,4(mol)$

$\to m_{KMnO_4(dùng)}=\frac{0,4.158}{80\%}=79(g)$

a, \(n_{Fe}=\dfrac{16,8}{56}=0,3\left(mol\right)\)

PTHH: 3Fe + 2O₂ ---t^o→ Fe₃O₄

Mol:     0,3      0,2               0,1

\(m_{Fe_3O_4}=0,1.232=23,2\left(g\right)\)

b, \(V_{O_2}=0,2.22,4=4,48\left(l\right)\Rightarrow V_{kk}=4,48.5=22,4\left(l\right)\)

c, 

PTHH: 2KMnO₄ ---t^o→ K₂MnO₄ + MnO₂ + O₂

Mol:        0,4                                                 0,2

\(m_{KMnO_4\left(lt\right)}=0,4.158=63,2\left(g\right)\)

 \(\Rightarrow m_{KMnO_4\left(tt\right)}=\dfrac{63,2}{80\%}=79\left(g\right)\)

a)

\(4P + 5O_2 \xrightarrow{t^o} 2P_2O_5\)

Sản phẩm : Điphotpho pentaoxit.

b)

\(n_P = \dfrac{6,2}{31} = 0,2(mol)\\ \Rightarrow n_{P_2O_5} = \dfrac{1}{2}n_P = 0,1(mol)\\ \Rightarrow m_{P_2O_5} = 0,1.142 = 14,2(gam)\)

c)

\(n_{O_2} = \dfrac{5}{4}n_P = 0,125(mol)\\ \Rightarrow V_{O_2} = 0,125.22,4 = 2,8(lít)\)

d)

\(V_{không\ khí} = \dfrac{2,8}{20\%} = 14(lít)\)

Giúp mình với mình đang cần gấp😇

Số mol của 16.8g Fe 

n_Fe = \(\dfrac{m}{M}\) \(\dfrac{16.8}{56}\) = 0.3 mol

PTHH: 4Fe + 3O₂ \(\rightarrow\) 2Fe₂O₃

Tỉ lệ:      4        3               2

Mol:    0.3 \(\rightarrow\)   0.225    \(\rightarrow\)  0.15

a.Khổi lượng sản phẩm thu được:

m_Fe2O3 = n . M = 0.15 . 160 = 24g

b.Thể tích của O₂ ở đktc:

V_O2 = n . 24 = 0.225 . 22.4 = 5.04 l 

Thể tích của Không khí đã dùng

V_kk = \(\dfrac{5.04.100\%}{20\%}\) = 25.2 l

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

b)

\(n_{CH_4} = \dfrac{6,72}{22,4} = 0,3(mol)\\ n_{CO_2} = n_{CH_4} = 0,3(mol)\\ n_{H_2O} = 2n_{CH_4} = 0,6(mol)\\ \Rightarrow m_{sản\ phẩm} = 0,3.44 + 0,6.18 = 24(gam)\)

c)

\(n_{O_2} = 2n_{CH_4} = 0,3.2 = 0,6(mol)\\ \Rightarrow V_{O_2} = 0,6.22,4= 13,44(lít)\\ \Rightarrow V_{không\ khí} = 5V_{O_2} = 13.44.5 = 67,2(lít)\)

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

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

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

Theo gt ta có: $n_{H_2}=0,75(mol)$

a, $2H_2+O_2\rightarrow 2H_2O$

Ta có: $n_{O_2}=0,5.n_{H_2}=0,375(mol)\Rightarrow V_{O_2}=8,4(l)\Rightarrow V_{kk}=42(l)$

b, $2KMnO_4\rightarrow K_2MnO_4+MnO_2+O_2$

Ta có: $n_{KMnO_4}=2.n_{O_2}=0,75(mol)\Rightarrow m_{KMnO_4}=118,5(g)$

a)

\(2H_2 + O_2 \xrightarrow{t^o} 2H_2O\\ V_{O_2} = \dfrac{V_{H_2}}{2} = \dfrac{16,8}{2} = 8,4(lít)\\ V_{không\ khí} = \dfrac{8,4}{20\%} = 42(lít)\)

b)

\(n_{O_2} = \dfrac{8,4}{22,4} = 0,375(mol)\\ 2KMnO_4 \xrightarrow{t^o} K_2MnO_4 + MnO_2 + O_2\\ n_{KMnO_4} = 2n_{O_2} = 0,75(mol)\\ \Rightarrow m_{KMnO_4} = 0,75.158 = 118,5(gam)\\ 2KClO_3 \xrightarrow{t^o} 2KCl + 3O_2\\ n_{KClO_3} = \dfrac{2}{3}n_{O_2} = 0,25(mol)\\ \Rightarrow m_{KClO_3} = 0,25.122,5 = 30,625(gam)\)

\(n_{H_2}=\dfrac{16.8}{22.4}=0.75\left(mol\right)\)

\(2H_2+O_2\underrightarrow{t^0}2H_2O\)

\(0.75...0.375\)

\(V_{O_2}=0.375\cdot22.4=8.4\left(l\right)\)

\(V_{kk}=5V_{O_2}=8.4\cdot5=42\left(l\right)\)

\(2KMnO_4\underrightarrow{t^0}K_2MnO_4+MnO_2+O_2\)

\(0.75..............................................0.375\)

\(m_{KMnO_4}=0.75\cdot158=118.5\left(g\right)\)

\(2KClO_3\underrightarrow{t^0}2KCl+3O_2\)

\(0.25.......................0.375\)

\(m_{KClO_3}=0.25\cdot122.5=30.625\left(g\right)\)

\(n_{Fe}=\dfrac{42}{56}=0,75\left(mol\right)\\ a,PTHH:3Fe+2O_2\rightarrow\left(t^o\right)Fe_3O_4\\ b,n_{O_2}=\dfrac{2}{3}.0,75=0,5\left(mol\right)\\ V_{O_2\left(đktc\right)}=0,5.22,4=11,2\left(l\right)\\ V_{kk}=5.V_{O_2\left(đktc\right)}=5.11,2=56\left(l\right)\)

3Fe+2O₂-to->Fe₃O₄

0,75---0,5-- mol

n _Fe=\(\dfrac{42}{56}\)=0,75 mol

=>V_O2=0,5.22,4=11,2l

=>V_kk=11,2.5=56l