\(Fe_2O_3+3H_2\rightarrow\left(t^o\right)2Fe+3H_2O\\ CuO+H_2\rightarrow\left(t^o\right)Cu+H_2O\\ Đặt:n_{Fe}=a\left(mol\right);n_{Cu}=0,5a\left(mol\right)\\ m_{hhB}=17,6\\ \Leftrightarrow56a+64.0,5a=17,6\\ \Leftrightarrow a=0,2\left(mol\right)\\ \Rightarrow n_{Fe}=0,2\left(mol\right);n_{Cu}=0,1\left(mol\right)\\ a,n_{H_2}=\dfrac{3}{2}.n_{Fe}+n_{Cu}=\dfrac{3}{2}.0,2+0,1=0,4\left(mol\right)\\ \Rightarrow V_{H_2\left(đktc\right)}=0,4.22,4=8,96\left(l\right)\\ b,Fe+2HCl\rightarrow FeCl_2+H_2\\ n_{HCl}=\dfrac{18,25}{36,5}=0,5\left(mol\right)\\ Vì:\dfrac{0,2}{1}< \dfrac{0,5}{2}\Rightarrow HCldư\\ \Rightarrow ddC:FeCl_2,HCldư\\ n_{FeCl_2}=n_{Fe}=0,2\left(mol\right)\Rightarrow m_{FeCl_2}=0,2.127=25,4\left(g\right)\\ n_{HCl\left(dư\right)}=0,5-0,2.2=0,1\left(mol\right)\)

\(\Rightarrow m_{HCl\left(dư\right)}=0,1.36,5=3,65\left(g\right)\)

Bạn ghi lại CTHH đúng của "ngo, ceo, cup" nhé.

Bài 1 : 

a)
$Fe_2O_3 + 3H_2 \xrightarrow{t^o} 2F e+ 3H_2O$
$Fe_3O_4 + 4H_2 \xrightarrow{t^o} 3Fe + 4H_2O$

b)

Cách 1 : Gọi $n_{Fe_2O_3} = a ; n_{Fe_3O_4} = b$

Ta có :

$\dfrac{16(3a + 4b)}{160a + 232b}.100\% = 28,205\%$

$n_{H_2} = 3a + 4b = 2,2 : 2 = 1,1$

Suy ra: $a = 0,1 ; b = 0,2$

Suy ra: $m =0,1.160 + 0,2.232 = 62,4(gam)$

Cách 2 :

$n_{O(oxit)} = n_{H_2} = 1,1(mol)$
$m_O = 1,1.16 = 17,6(gam)$
$\Rightarrow m = 17,6 : 28,205\% = 62,4(gam)$

c)

$m_{Fe_2O_3} = 0,1.160 = 16(gam)$
$m_{Fe_3O_4} = 0,2.232 = 46,4(gam)$
d)

$n_{Fe} = 2a + 3b = 0,8(mol)$
$m_{Fe} = 0,8.56 = 44,8(gam)$

Bài 2 : 

\(Ta.có:\dfrac{n_{Mg}}{n_{Zn}}=\dfrac{3}{1}\Rightarrow\dfrac{\dfrac{m_{Mg}}{M_{Mg}}}{\dfrac{m_{Zn}}{M_{Zn}}}=\dfrac{m_{Mg}}{24}.\dfrac{65}{m_{Zn}}=\dfrac{3}{1}\Rightarrow\dfrac{m_{Mg}}{m_{Zn}}=\dfrac{3}{1}:\dfrac{65}{24}=\dfrac{72}{24}\)

\(\Rightarrow n_{Mg}=13,7:\left(72+24\right).3=0,3\left(mol\right)\\ \Rightarrow n_{Zn}=\dfrac{n_{Mg}}{3}=\dfrac{0,3}{3}=0,1\left(mol\right)\)

\(\Rightarrow m_{Mg}=n.M=0,3.24=7,2\left(g\right)\\ \Rightarrow m_{Zn}=n.M=0,12.65=6,5\left(g\right)\)

\(a,PTHH:2Mg+O_2\underrightarrow{t^o}2MgO\left(1\right)\\ PTHH:2Zn+O_2\underrightarrow{t^o}2ZnO\left(2\right)\)

\(Theo.PTHH\left(1\right):n_{O_2\left(1\right)}=\dfrac{1}{2}.n_{Mg}=\dfrac{1}{2}.0,3=0,15\left(mol\right)\\ Theo.PTHH\left(2\right):n_{O_2\left(2\right)}=\dfrac{1}{2}.n_{Zn}=\dfrac{1}{2}.0,1=0,05\left(mol\right)\\ n_{O_2\left(tổng\right)}=n_{O_2\left(1\right)}+n_{O_2\left(2\right)}=0,15+0,05=0,2\left(mol\right)\\ V_{O_2\left(tổng,đktc\right)}=n.22,4=0,2.22,4=4,48\left(l\right)\)

\(b,Theo.PTHH\left(1\right):n_{MgO}=n_{Mg}=0,3\left(mol\right)\\ m_{MgO}=n.M=0,3.40=12\left(g\right)\\ Theo.PTHH\left(2\right):n_{ZnO}=n_{Zn}=0,1\left(mol\right)\\ m_{ZnO}=n.M=0,1.81=8,1\left(g\right)\\ m=m_{hh}=m_{MgO}+m_{ZnO}=12+8,1=20,1\left(g\right)\)

\(n_{CuO}=2a\left(mol\right)\Rightarrow n_{Fe_2O_3}=a\left(mol\right)\)

\(m_X=80\cdot2a+160a=80\left(g\right)\)

\(\Rightarrow a=0.25\left(mol\right)\)

\(CuO+H_2\underrightarrow{^{^{t^0}}}Cu+H_2O\)

\(Fe_2O_3+3H_2\underrightarrow{^{^{t^0}}}2Fe+3H_2O\)

\(n_{H_2}=0.5+0.25\cdot3=1.25\left(mol\right)\)

\(V_{H_2}=1.25\cdot22.4=28\left(l\right)\)

\(m_{cr}=0.5\cdot64+0.5\cdot56=60\left(g\right)\)

a)

\(\left\{{}\begin{matrix}n_{Fe_3O_4}=3a\left(mol\right)\\n_{CuO}=2a\left(mol\right)\end{matrix}\right.\left(a,b>0\right)\\ m_{hh}=85,6\\ \Leftrightarrow232.3a+80.2a=85,6\\ \Leftrightarrow a=0,1\\ \Rightarrow n_{Fe_3O_4}=3a=3.0,1=0,3\left(mol\right)\\n_{CuO}=2.0,1=0,2\left(mol\right)\\ m_{Fe}=n_{Fe}.M_{Fe} =0,3.3.56=50,4\left(g\right)\\ m_{Cu}=n_{Cu}.M_{Cu}=0,1.64=6,4\left(g\right)\)

b) Sao lại có khí H₂ ở đây em nhỉ?

\(\left\{{}\begin{matrix}CuO:a\\Fe2O3:2a\end{matrix}\right.\)

a.\(80a+320a=24\Leftrightarrow a=0.06\)

\(\Rightarrow\left\{{}\begin{matrix}CuO=0.06\\Fe2O3=0.12\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}CuO=4.8g\\Fe2O3=19.2g\end{matrix}\right.\)

b.\(CuO+H2\rightarrow Cu+H2O\)

    a           a         a

\(Fe2O3+3H2\rightarrow2Fe+3H2O\)

    2a          6a       4a

\(\Rightarrow V_{H2}=\left(a+6a\right)\times22.4=9.408l\)

c.nHCl = 0.2 mol

\(Fe+2HCl\rightarrow FeCl2+H2\)

 0.1     0.2

m chất rắn còn lại = mCu + m Fe ban đầu - m Fe bị hòa tan

                             = \(a\times64+4a\times56-0.1\times56=11.68g\)

\(n_{Pb}=\dfrac{2,07}{207}=0,01mol\)

\(n_{Cu}=\dfrac{1,6}{64}=0,025mol\)

\(PbO+H_2\rightarrow\left(t^o\right)Pb+H_2O\)

0,01    0,01            0,01            ( mol )

\(CuO+H_2\rightarrow\left(t^o\right)Cu+H_2O\)

0,025  0,025        0,025            ( mol )

\(m_{hh}=m_{PbO}+m_{CuO}=\left(0,01.223\right)+\left(0,025.80\right)=4,23g\)

\(V_{H_2}=\left(0,01+0,025\right).22,4=0,784l\)

\(n_{Pb}=\dfrac{2,07}{207}=0,01\left(mol\right)\\ n_{Cu}=\dfrac{1,6}{64}=0,025\left(mol\right)\\ PTHH:PbO+H_2\underrightarrow{t^o}Pb+H_2O\\ Mol:0,01\leftarrow0,01\leftarrow0,01\\ CuO+H_2\underrightarrow{t^o}Cu+H_2O\\ Mol:0,025\leftarrow0,025\leftarrow0,025\)

\(\Rightarrow\left\{{}\begin{matrix}m_{CuO}=0,025.80=2\left(g\right)\\m_{PbO}=0,01.223=2,23\left(g\right)\end{matrix}\right.\Rightarrow m_{oxit}=2+2,23=4,23\left(g\right)\\ V_{H_2}=\left(0,01+0,025\right).22,4=0,784\left(l\right)\)

Quy đổi E thành:

C 2 H 3 O N :   0 , 22   m o l (Tính từ  n N 2   =   0 , 11 )

C H 2 :   a   m o l H 2 O :   b   m o l

C 2 H 3 O N   +   2 , 25 O 2   →   2 C O 2   +   1 , 5 H 2 O   +   0 , 5 N 2 C H 2 + 1 , 5 O 2   →   C O 2   +   H 2 O = >   n O 2   =   2 , 25   .   0 , 22   +   1 , 5 a   =   0 , 99

=> a = 0,33

n C O 2   =   a   +   0 , 44 n H 2 O   =   a   +   b   +   0 , 33

=> 44(a + 0,44) + 18(a + b + 0,33) = 46,48

=> b = 0,04

Đặt x, 3x là số mol X, Y

=> x + 3x = b = 0,04 => x = 0,01

Đặt m, n là số gốc amino axit trong X và Y

=> Số -CONH- = m + n- 2 = 8

1 mắt xích X và Y chứa 1 N => bảo toàn N ta có:  n N   ( t r o n g   X   +   Y )   =   2. n N 2   s i n h   r a

=> 0,01m + 0,03n = 0,22

=> m = 4 và n = 6

Đặt u, v là số mol của Gly và Val

= >   n N   =   u   +   v   =   0 , 22 v à   n C   =   2 u   +   5 v   =   a   +   0 , 44   =   0 , 77

=> u = v = 0,22

=> Thủy phân hoàn toàn hỗn hợp E thu được  n G l y   :   n A l a   =   1   :   1

Đáp án cần chọn là: B