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a, \(n_{H_2}=\dfrac{0,896}{22,4}=0,04\left(mol\right)\)
PTHH: Fe + 2HCl ---> FeCl2 + H2
0,04<-----------------------0,04
\(m_{Cu}=3,52-0,04.56=1,28\left(g\right)\)
Bảo toàn O: \(\left\{{}\begin{matrix}n_{O\left(oxit\right)}=\dfrac{4,8-3,52}{16}=0,08\left(mol\right)\\n_{O\left(CuO\right)}=n_{Cu}=\dfrac{1,28}{64}=0,02\left(mol\right)\end{matrix}\right.\)
=> \(n_{O\left(Fe_xO_y\right)}=0,08-0,02=0,06\left(mol\right)\)
PTHH:CuO + H2 --to--> Cu + H2O
0,02<--------------0,02
=> \(\left\{{}\begin{matrix}\%m_{CuO}=\dfrac{0,02.80}{4,8}.100\%=33,33\%\\\%m_{Fe_xO_y}=100\%-33,33\%=66,67\%\end{matrix}\right.\)
b, CTHH là FexOy
=> x : y = 0,04 : 0,06 = 2 : 3
=> CTHH là Fe2O3
- Cho phản ứng xảy ra hoàn toàn (2 chất trong A có sắt và oxit khác oxit sắt ban đầu)
\(yH_2+Fe_xO_y\rightarrow\left(t^o\right)xFe+yH_2O\left(1\right)\\ Fe+2HCl\rightarrow FeCl_2+H_2\left(2\right)\\ n_{H_2\left(2\right)}=n_{Fe\left(2\right)}=n_{Fe\left(1\right)}=0,3\left(mol\right)\\ n_{O\left(trong.oxit\right)}=n_{H_2O}=n_{H_2}=0,4\left(mol\right)\\ BTKL:m_{H_2}+m_{oxit}=m_A+m_{H_2O}\\ \Leftrightarrow0,4.2+m=28,4+18.0,4\\ \Leftrightarrow m=34,8\left(g\right)\\ b,x:y=0,3:0,4=3:4\Rightarrow x=3;y=4\\ \Rightarrow CTHH:Fe_3O_4\)
PTHH: \(Fe_xO_y+yH_2\underrightarrow{t^o}xFe+yH_2O\)
\(Fe+2HCl\rightarrow FeCl_2+H_2\uparrow\)
Đặt \(\left\{{}\begin{matrix}n_{Fe\left(oxit\right)}=a\left(mol\right)=n_{H_2}\\n_{O\left(oxit\right)}=b\left(mol\right)\end{matrix}\right.\)
Ta có: \(m_{tăng}=m_{Fe}-m_{H_2}\) \(\Rightarrow56a-2a=3,24\) \(\Rightarrow a=n_{Fe}=0,06\left(mol\right)\)
Hỗn hợp D gồm \(\left\{{}\begin{matrix}n_{CO_2\left(dư\right)}=c\left(mol\right)\\n_{H_2O}=n_{O\left(oxit\right)}=b\left(mol\right)\end{matrix}\right.\)
Ta có hệ phương trình: \(\left\{{}\begin{matrix}c+b=0,1\\18b+2c=7,4\cdot2\cdot\left(b+c\right)\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}b=0,08\\c=0,02\end{matrix}\right.\)
\(\Rightarrow x:y=a:b=0,06:0,08=3:4\)
\(\Rightarrow\) Công thức cần tìm là Fe3O4
Gọi CT oxit sắt là FexOy
Gọi nCu=a(mol)
nH2=\(\dfrac{6,72}{22,4}\)=0,3(mol)
FexOy+yH2to→xFe+yH2O(1)
Fe+2HCl→FeCl2+H2(2)
Theo pthh(2)
nFe=nH2=0,3(mol)
Theo pthh(1)
nFexOy=\(\dfrac{0,3}{x}\)(mol)
Ta có: 64a+56.0,3=29,6
⇒a=0,2(mol)
⇒mCu=0,2.64=12,8(g)
⇒mFexOy=36−12,8=23,2(g)
=>MFexOy= \(\dfrac{\dfrac{23,2}{0,3}}{x}\)=\(\dfrac{232x}{3}\)
=>56x+16y=\(\dfrac{232x}{3}\)
=>\(\dfrac{64x}{3}=16y\)
->\(\dfrac{x}{y}=\dfrac{3}{4}\)
⇒CTHH:Fe3O4
Ta có :
%m Cu=\(\dfrac{12,8}{36}100\)=35,56%
=>%m Fe3O4=100%-35,56%=64,44%
a) PTHH : \(FeO+H_2-t^o->Fe+H_2O\)
\(CuO+H_2-t^o->Cu+H_2O\)
Đặt \(\hept{\begin{cases}n_{FeO}=x\left(mol\right)\\n_{CuO}=y\left(mol\right)\end{cases}}\) => \(72x+80y=11,2\left(I\right)\)
Có : \(m_{O\left(lấy.đi\right)}=m_{giảm}=1,92\left(g\right)\)
=> \(n_{O\left(lấy.đi\right)}=\frac{1,92}{16}=0,12\left(mol\right)\) Vì H% = 80% => Thực tế : \(n_{O\left(hh\right)}=\frac{0,12}{80}\cdot100=0,15\left(mol\right)\)
BT Oxi : \(x+y=0,15\left(II\right)\)
Từ (I) và (II) suy ra : \(\hept{\begin{cases}x=0,1\\y=0,05\end{cases}}\)
=> \(\hept{\begin{cases}m_{FeO}=7,2\left(g\right)\\m_{CuO}=4\left(g\right)\end{cases}}\)
b) PTHH : \(Fe+H_2SO_4-->FeSO_4+H_2\)
BT Fe : \(n_{Fe}=n_{FeO}=0,1\left(mol\right)\)
Theo pthh : \(n_{H_2}=n_{Fe}=0,1\left(mol\right)\)
=> \(V_{H_2}=2,24\left(l\right)\)
BT Cu : \(n_{Cu}=n_{CuO}=0,05\left(mol\right)\)
=> \(m_{CR\left(ko.tan\right)}=0,05\cdot64=3,2\left(g\right)\)
nH2= 0,448/22,4= 0,02(mol)
PTHH :
CuO + H2 -tdo--> Cu + H20
FexOy + yH2 -tdo-> xFe + yH20
Cu + HCl --> k pu
Fe + 2HCl ---> FeCl2 + H2
0,02 -- 0,04---> 0,02 --- 0,02 (mol)
mFe = 0,02 .56= 1,12(g)
=> mCu = 1,76 - 1,12= 0,64(g)
n Cu = 0,64 /64 =0,01(mol)
PTHH :
CuO + H2 -tdo-> Cu + H20
0,,01 --0,01 ----> 0,01(mol)
mCuO= 0,01 . 80 = 0,8(g)
=> mFexOy = 2,4-0,8= 1,6(g)
PTHH :
FexOy + yH2 ---> xFe + yH20
56x+ 16y ---------> 56x
1,6 (g) -------------> 1,12(g)
<=> 1,6 .56x = 1,12( 56x + 16y)
<=> 89,6x = 62,72 x + 17,92y
<=> 89,6x - 62,72x = 17,92y
<=> 26,88 x = 17,92y
=> x/y= 17,92 / 26,88 =2/3
Vậy công thức đúng là Fe203.
\(n_{H_2}=\dfrac{4,032}{22,4}=0,18\left(mol\right)\)
PTHH: Fe + H2SO4 ---> FeSO4 + H2
0,18 <------------------------ 0,18
\(\rightarrow n_O=\dfrac{13,92-0,18.56}{16}=0,24\left(mol\right)\)
CTHH: FexOy
=> x : y = 0,18 : 0,24 = 3 : 4
CTHH Fe3O4
a) A gồm Cu, Fe
\(n_O=\dfrac{39,2-29,6}{16}=0,6\left(mol\right)\)
=> \(n_{H_2O}=0,6\left(mol\right)\)
=> \(n_{H_2}=0,6\left(mol\right)\)
=> \(V_{H_2}=0,6.22,4=13,44\left(l\right)\)
b)
Gọi \(\left\{{}\begin{matrix}n_{CuO}=a\left(mol\right)\\n_{Fe_xO_y}=b\left(mol\right)\end{matrix}\right.\)
=> 80a + b(56x + 16y) = 39,2
=> 80a + 56bx + 16by = 39,2 (1)
nO = 0,6 (mol)
=> a + by = 0,6
=> 80a + 80by = 48 (2)
\(n_{H_2}=\dfrac{6,72}{22,4}=0,3\left(mol\right)\)
PTHH: Fe + 2HCl --> FeCl2 + H2
0,3<-------------------0,3
=> nFe = bx = 0,3 (mol)
(2) - (1) => 64by - 56bx = 8,8
=> by = 0,4
Xét \(\dfrac{bx}{by}=\dfrac{x}{y}=\dfrac{0,3}{0,4}=\dfrac{3}{4}\)
=> CTHH: Fe3O4
Có: \(\left\{{}\begin{matrix}80a+232b=39,2\\a+4b=0,6\end{matrix}\right.\)
=> a = 0,2; b = 0,1
=> \(\left\{{}\begin{matrix}m_{CuO}=0,2.80=16\left(g\right)\\m_{Fe_3O_4}=0,1.232=23,2\left(g\right)\end{matrix}\right.\)