\(n_{Na}=\dfrac{4.6}{23}=0.2\left(mol\right)\)

\(n_K=\dfrac{3.9}{39}=0.1\left(mol\right)\)

\(Na+H_2O\rightarrow NaOH+\dfrac{1}{2}H_2\)

\(K+H_2O\rightarrow KOH+\dfrac{1}{2}H_2\)

\(V_{H_2}=\left(\dfrac{0.2}{2}+\dfrac{0.1}{2}\right)\cdot22.4=3.36\left(l\right)\)

\(m_{bazo}=0.2\cdot40+0.1\cdot56=13.6\left(g\right)\)

\(n_{Fe}=\dfrac{1,68}{56}=0,03\left(mol\right)\)

\(3Fe+2O_2\rightarrow\left(t^o\right)Fe_3O_4\)

 0,03   0,02               0,01  ( mol )

\(m_{Fe_3O_4}=0,01.232=2,32\left(g\right)\)

\(V_{kk}=0,02.22,4.5=2,24\left(l\right)\)

\(2KClO_3\rightarrow\left(t^o\right)2KCl+3O_2\)

   \(\dfrac{1}{75}\)                               0,02   ( mol )

\(m_{KClO_3}=\dfrac{1}{75}.122,5=1,63\left(g\right)\)

a,nFe=1,68/56=0,03 mol

Ta có PTHH : 3Fe + 2O2 --> Fe3O4 (1) ( ở trên dấu --> có to nha )

Theo PTHH ta có :

nFe3O4=1/3nFe=1/3.0,03=0,01 mol

nO2=2/3nFe=2/3.0,03=0,02 mol

=>mFe3O4= 0,01.232=2,32g

=>Vkk=5.(0,02.22,4)=2,24 l

b, Ta có PTHH: 2KClO3 --> 2KCl + 3O2 (2) ( trên dấu --> vẫn có to )

Gọi x là số mol KClO3 cần dùng ( x > 0 )

Theo PTHH (3) và theo bài ra ta có PTHH sau:

2/3x=0,02 

=> x=0,03 mol

=> mKClO3= 0,03.122,5= 3,675g

Bài 1:

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

Ta có: \(n_{Al}=\dfrac{10,8}{27}=0,4\left(mol\right)\)

\(\Rightarrow\left\{{}\begin{matrix}n_{O_2}=0,3mol\\n_{Al_2O_3}=0,2mol\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}V_{O_2}=0.3\cdot22.4=6,72\left(l\right)\\m_{Al_2O_3}=0,2\cdot102=20,4\left(g\right)\end{matrix}\right.\)

Bài 2 : 

\(n_{Na_2O} = \dfrac{12,4}{62} = 0,2(mol)\)

4Na + O₂ \(\xrightarrow{t^o}\) 2Na₂O

0,4......0,1.........0,2..................(mol)

Vậy  :

\(V_{O_2} = 0,1.22,4 = 2,24(lít)\\ m_{Na} = 0,4.23 = 9,2(gam)\)

Bài 1:

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

Ta có: \(n_{Al}=\dfrac{10,8}{27}=0,4\left(mol\right)\)

Theo PT: \(\left\{{}\begin{matrix}n_{O_2}=\dfrac{3}{4}n_{Al}=0,3\left(mol\right)\\n_{Al_2O_3}=\dfrac{1}{2}n_{Al}=0,2\left(mol\right)\end{matrix}\right.\)

b, Ta có: \(V_{O_2}=0,3.22,4=6,72\left(l\right)\)

c, Ta có: \(m_{Al_2O_3}=0,2.102=20,4\left(g\right)\)

Bài 2:

PT: \(4Na+O_2\underrightarrow{t^o}2Na_2O\)

Ta có: \(n_{Na_2O}=\dfrac{12,4}{62}=0,2\left(mol\right)\)

Theo PT: \(\left\{{}\begin{matrix}n_{Na}=2n_{Na_2O}=0,4\left(mol\right)\\n_{O_2}=\dfrac{1}{2}n_{Na_2O}=0,1\left(mol\right)\end{matrix}\right.\)

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

b, Ta cóL \(m_{Na}=0,4.23=9,2\left(g\right)\)

Bạn tham khảo nhé!

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)

$\rm a)n_{kk} = \dfrac{67,2}{22,4} = 3 (mol)$
$\rm \Rightarrow n_{O_2} = 20\%.3 = 0,6 (mol)$
$\rm n_P = \dfrac{24,8}{31} = 0,8 (mol)$
PTHH:        \(\rm 4P + 5O_2 \xrightarrow{t^o} 2P_2O_5 \)

Ban đầu:  0,8      0,6

Pư:          0,48<--0,6

Sau pư:   0,32      0          0,24

$\rm \Rightarrow m_{\text{sản phẩm tạo thành}} = m_{P_2O_5(sinh.ra)} = 0,24.142 = 34,08 (g)$

$\m b) m_{hh} = m_{P(dư)} + m_{P_2O_5} = 0,32.31 + 34,08 = 44 (g)$
$\rm \Rightarrow \%m_P = \dfrac{0,32.31}{44} .100\% = 22,545\%$
$\rm \Rightarrow \%m_{P_2O_5} = 100\% - 22,545\% = 77,455\%$

\(n_P=\dfrac{24,8}{31}=0,8\left(mol\right)\)

Thể tích Oxi trong 67,2 lít không khí : 

67,2 x 20% = 13,44(l) 

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

PTHH : 

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

Trc p/ư:   0,8     0,6                    (mol) 

p/ư         0,48    0,6      0,24

Sau p/ư: 0,32   0          0,24  

=> Sau p/ư P dư 

Khối lượng sản phẩm tạo thành :

\(m_{P_2O_5}=0,24.142=34,08\left(g\right)\)

Khối lượng P trong hỗn hợp : 

\(m_{P\left(P_2O_5\right)}=0,48.31=14,88\left(g\right)\)

Thành phần % của P : 

\(14,88:34,08=43,66\%\)

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

Ta có: \(\left\{{}\begin{matrix}n_{Fe}=\dfrac{11,2}{56}=0,2\left(mol\right)\\n_{O_2}=\dfrac{12,8}{32}=0,4\left(mol\right)\end{matrix}\right.\)

Xét tỉ lệ: \(\dfrac{0,2}{3}< \dfrac{0,4}{2}\) \(\Rightarrow\) Oxi còn dư, Fe p/ứ hết

\(\Rightarrow n_{O_2\left(dư\right)}=0,4-\dfrac{2}{15}=\dfrac{4}{15}\left(mol\right)\)

+) Theo PTHH: \(\left\{{}\begin{matrix}n_{O_2}=\dfrac{2}{15}\left(mol\right)\\n_{Fe_3O_4}=\dfrac{1}{15}\left(mol\right)\end{matrix}\right.\)

\(\Rightarrow\left\{{}\begin{matrix}V_{kk}=\dfrac{2}{15}\cdot22,4\cdot5\approx14,93\left(l\right)\\m_{Fe_3O_4}=\dfrac{1}{15}\cdot232\approx15,47\left(g\right)\end{matrix}\right.\) 

a) \(n_{C_2H_4}=\dfrac{4,958}{24,79}=0,2\left(mol\right)\)

PTHH: \(C_2H_4+3O_2\xrightarrow[]{t^o}2CO_2+2H_2O\)

              0,2---->0,6---->0,4---->0,4

\(\Rightarrow V_{O_2}=0,6.24,79=14,874\left(l\right)\)

b) \(V_{CO_2}=0,4.22,4=9,916\left(l\right)\)

c) \(Ca\left(OH\right)_2+CO_2\rightarrow CaCO_3\downarrow+H_2O\)

                       0,4------>0,4

\(\Rightarrow\left\{{}\begin{matrix}x=m_{CaCO_3}=0,4.100=40\left(g\right)\\y=m_{b\text{ình}.t\text{ăng}}=m_{CO_2}+m_{H_2O}=0,4.44+0,4.18=24,8\left(g\right)\end{matrix}\right.\)