nO2(tổng)=0,3(mol); nCO2=0,2(mol)

CO + 1/2 O2 -to-> CO2 (1)

0,2<----0,1<-------0,2(mol)

H2 + 1/2 O2 -to-> H2O (2)

0,4<---0,2<-------0,4(mol)

nO2(2)= nO2(tổng)- nO2(1)=0,3-0,1=0,2(mol)

Vì số mol tỉ lệ thuận thể tích:

=> %V(CO/hh)= [0,2/(0.2+0,4)].100=33,333%

=>%V(H2/hh)=100%-33,333%=66,667%

\(n_{O_2}=\dfrac{6,72}{22,4}=0,3\left(mol\right)\\ n_{CO_2}=\dfrac{4,48}{22,4}=0,2\left(mol\right)\)

PTHH:

2CO + O₂ --t^o--> 2CO₂

0,2<---0,1<--------0,2

2H₂ + O₂ --t^o--> 2H₂O

0,4<--0,2<-------0,2

\(\Rightarrow\left\{{}\begin{matrix}\%V_{CO}=\dfrac{0,2}{0,2+0,4}.100\%=33,33\%\\\%V_{H_2}=100\%-33,33\%=66,67\%\end{matrix}\right.\)

\(n_{CO_2}=\dfrac{6,72}{22,4}=0,3\left(mol\right)\\ n_{O_2}=\dfrac{4,48}{22,4}=0,2\left(mol\right)\)

PTHH: 2CO + O₂ --t^o--> 2CO₂

            0,3<--0,15<------0,3

            2H₂ + O₂ --t^o--> 2H₂O

            0,1<--0,05

\(\Rightarrow\left\{{}\begin{matrix}\%V_{CO}=\%n_{CO}=\dfrac{0,3}{0,3+0,1}.100\%=75\%\\\%V_{H_2}=100\%-75\%=25\%\end{matrix}\right.\)

\(n_{CO_2}=\dfrac{4,958}{24,79}=0,2\left(mol\right)\\ n_{O_2}=\dfrac{7,437}{24,79}=0,3\left(mol\right)\)

PTHH:

2CO + O₂ --to--> 2CO₂

0,2       0,1             0,2

-> n_O2 = 0,3 - 0,1 = 0,2 (mol)

2H₂ + O₂ --to--> 2H₂O

0,4      0,2

\(\rightarrow n_{hhkhí}=0,1+0,4=0,5\left(mol\right)\\ \rightarrow\left\{{}\begin{matrix}\%V_{CO}=\dfrac{0,2}{0,5}=40\%\\\%V_{H_2}=100\%-40\%=60\%\end{matrix}\right.\)

n CO2 = 0,2 mol
n O2 = 0,3 mol
2 CO + O2 -> 2 CO2
............0,1........0,2
2 H2 + O2 -> 2 H2O
.......0,3-0,1.....0,4
=> % tự tính nha

16 nCO2=0,2mol

PTHH: 2CO+O2=>2CO2

         0,2<--0,1<---0,2

=> mO2=0,2.32=6,4g

=> khối lượng Oxi phản ứng với H2 là :

9,6-6,4=3,2g

=> nH2O=3,2:32=0,1mol

PTHH: 2H+O2=>H2O

b)

           0,2<-0,1<-0,2

=> mH2=2.0,2=0,4g

mCO =0,2.28=5,6g

=> m hh=5,6+0,4=6g

CuO+H2-to--->Cu+H2O 

0,6----0,6
nCuO =48/80=0,6 (mol)
==>VH2 =0,6×22,4=13.44(l)

17.

\(n_{Fe}=\dfrac{5,6}{56}=0,1mol\)

\(m_{H_2SO_4}=200.19,6\%=39,2g\)

\(n_{H_2SO_4}=\dfrac{39,2}{98}=0,4mol\)

\(Fe+H_2SO_4\rightarrow FeSO_4+H_2\)

0,1 <   0,4                                 ( mol )

0,1       0,1              0,1        0,1   ( mol )

\(V_{H_2}=0,1.22,4=2,24l\)

Chất còn dư là H2SO4

\(m_{H_2SO_4\left(dư\right)}=\left(0,4-0,1\right).98=29,4g\)

\(\left\{{}\begin{matrix}m_{FeSO_4}=0,1.152=15,2g\\m_{H_2}=0,1.2=0,2g\end{matrix}\right.\)

\(m_{ddspứ}=5,6+200-0,1.2=205,4g\)

\(\left\{{}\begin{matrix}C\%_{FeSO_4}=\dfrac{15,2}{205,4}.100=7,4\%\\C\%_{H_2}=\dfrac{0,2}{205,4}.100=0,09\%\\C\%_{H_2SO_4\left(dư\right)}=\dfrac{29,4}{205,4}.100=14,31\%\end{matrix}\right.\)

\(n_{Fe}=\dfrac{5,6}{56}=0,1\left(mol\right)\)
\(m_{H_2SO_4}=\dfrac{200.19,6}{100}=39,2\left(g\right)\)
\(n_{H_2SO_4}=\dfrac{39,2}{98}=0,4\left(mol\right)\)
pthh : \(Fe+H_2SO_4->FeSO_4+H_2\)
 LTL: 
\(\dfrac{0,1}{1}< \dfrac{0,4}{1}\)
=> H2SO4 dư 
=> \(n_{H_2}=0,1mol\)
=> \(V_{H_2}=0,1.22,4=2,24\left(L\right)\)
theo pthh :\(n_{H_2SO_4\left(p\text{ư}\right)}=n_{Fe}=0,1\left(mol\right)\)
=> \(n_{H_2SO_4\left(d\right)}=0,4-0,1=0,3\left(mol\right)\)
=> \(m_{H_2SO_4}=98.0,3.29,4\left(g\right)\)
ta có mdd =5,6+39,2 = 44,8 (g)
\(C\%=\dfrac{5,6}{44,8}.100\%=12,55\)

\(n_{CO_2}=\dfrac{8.8}{44}=0.2\left(mol\right)\)

\(n_{O_2}=\dfrac{9.6}{32}=0.3\left(mol\right)\)

\(2CO+O_2\underrightarrow{t^0}2CO_2\)

\(0.2.......0.1.......0.2\)

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

\(0.4......0.3-0.1\)

\(\%m_{CO}=\dfrac{0.2\cdot28}{0.2\cdot28+0.4\cdot2}\cdot100\%=87.5\%\)

\(\%m_{H_2}=100-87.5=12.5\%\)

a)

\(2CO + O_2 \xrightarrow{t^o} 2CO_2(1)\\ 2H_2 + O_2 \xrightarrow{t^o} 2H_2O(2) \)

b)

\(n_{CO_2} = \dfrac{8,8}{44} = 0,2(mol)\\ n_{O_2} = \dfrac{9,6}{32} = 0,3(mol)\)

Theo PTHH :

\(n_{CO} = n_{CO_2} = 0,2(mol)\\ n_{O_2(1)} = \dfrac{1}{2}n_{CO_2} = 0,1(mol)\\ n_{H_2} = 2n_{O_2(2)} = 2(0,3-0,1) = 0,4(mol)\)

Vậy :

\(\%m_{CO} = \dfrac{0,2.28}{0,2.28+0,4.2}.100\% = 87,5\%\\ \%m_{H_2} = 100\% - 87,5\% = 12,5\%\)

a, Có: \(n_{O_2}=\dfrac{21,28}{22,4}=0,95\left(mol\right)\)

Theo ĐLBT KL, có: m + m_O2 = m_CO2 + m_H2O

⇒ m = 28,6 + 14,4 - 0,95.32 = 12,6 (g)

b, Có: \(n_{CO_2}=\dfrac{28,6}{44}=0,65\left(mol\right)\)

\(n_{H_2O}=\dfrac{14,4}{18}=0,8\left(mol\right)\)

BTNT O, có: n_CO + 2n_O2 = 2n_CO2 + n_H2O

⇒ n_CO = 0,65.2 + 0,8 - 0,95.2 = 0,2 (mol)

⇒ m_CO = 0,2.28 = 5,6 (g)

\(\Rightarrow\%m_{CO}=\dfrac{5,6}{12,6}.100\%\approx44,44\%\)

Bạn tham khảo nhé!

a) PTHH: \(2CO+O_2\underrightarrow{t^o}2CO_2\)  (1)

                \(4H_2+O_2\underrightarrow{t^o}2H_2O\)  (2)

b) Ta có: \(\left\{{}\begin{matrix}\Sigma n_{O_2}=\dfrac{9,6}{32}=0,3\left(mol\right)\\n_{CO_2}=\dfrac{8,8}{44}=0,2\left(mol\right)\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}n_{O_2\left(1\right)}=0,1mol\\n_{O_2\left(2\right)}=0,2mol\end{matrix}\right.\)

\(\Rightarrow\left\{{}\begin{matrix}m_{CO}=0,1\cdot28=2,8\left(g\right)\\m_{H_2}=0,2\cdot2=0,4\left(g\right)\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}\%m_{CO}=\dfrac{2,8}{2,8+0,4}\cdot100\%=87,5\%\\\%m_{H_2}=12,5\%\end{matrix}\right.\)

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

Theo PTHH: \(n_{KMnO_4}=2n_{O_2}=0,6mol\)

\(\Rightarrow m_{KMnO_4}=0,6\cdot158=94,8\left(g\right)\)