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2A+2aHCl->2ACla+aH2
2B+2bHCl->2BClb+aH2
nH2=0.3(mol)
->nHCl=0.3*2=0.6(mol)
->nCl/HCl=0.6(mol)
m muối khan=m kim loại+mCl/HCl=8+0.6*35.5=29.3(g)
\(2Al+3H_2SO_4\rightarrow Al_2\left(SO_4\right)_3+3H_2\)
x_______3/2x______________________
\(Fe+H_2SO_4\rightarrow FeSO_4+H_2\)
y ____ y___________________
Đổi 400ml = 0,4l
\(n_{H2SO4}=0,4.2=0,8\left(mol\right)\)
Giải hệ phương trình :
\(\left\{{}\begin{matrix}27x+56y=22\\\frac{3}{2}x+y=0,8\end{matrix}\right.\rightarrow\left\{{}\begin{matrix}x=0,4\\y=0,2\end{matrix}\right.\)
\(\rightarrow\%m_{Al}=\frac{27.0,4}{22}.100\%=49,1\%\)
\(\rightarrow\%m_{Fe}=100\%-49,1\%=50,9\%\)
\(n_{H2SO4}=n_{H2}=0,8\left(mol\right)\)
\(n_{Cl2}=\frac{13,44}{22,4}=0,6\left(mol\right)\)
\(PTHH:H_2+Cl_2\rightarrow2HCl\left(1\right)\)
Ban đầu :0,8____0,6__________(mol)
Phứng :0,6_____0,6_______1,2_(mol)
Sau phứng :0,2__0______1,2____(mol)
\(HCl+AgNO_3\rightarrow AgCl+HNO_3\left(2\right)\)
0,08______________0,08___________(mol)
\(n_{AgCl}=\frac{11,48}{108+35,5}=0,08\left(mol\right)\)
\(n_{HCl\left(1\right)_{thuc.te}}=0,08.\frac{470,8}{50}=0,75328\)
\(\rightarrow H=\frac{0,75328}{1,2}.100\%=62,73\%\)
PTHH: K2CO3 + 2 HCl ->2 KCl + H2O + CO2
x___________2x______2x____________x(mol)
KHCO3 + HCl -> KCl + H2O + CO2
y____y__________y_______y(mol)
mHCl= 27,375.0,2= 5,475
Ta có hpt:
\(\left\{{}\begin{matrix}2.36,5x+36,5y=5,475\\22,4x+22,4y=2,24\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=0,05\left(mol\right)\\x=0,05\left(mol\right)\end{matrix}\right.\)
mK2CO3= 0,05.138= 6,9(g)
mKHCO3= 0,05.100=5(g)
=> %mK2CO3= (6,9/11,9).100=57,893%
=> %mKHCO3= 100%- 57,893%= 42,107%
c) mKCl= 0,15. 74,5=11,175(g)
mddKCl= mhh+ mddHCl - mCO2= 11,9+27,375- 0,1.44=34,875(g)
=> C%ddKCl = (11,175/34,875).100=32,043%
vì cho B vào H2SO4 k thấy khí thoát ra => chất rắn B là Cu
Gọi nFe=a nCu(pư)=b
56a+64b=15,28-1,92=13,36 (1)
Fe + Fe2(SO4)3 ----> 3FeSO4
a--------a
Cu + Fe2(SO4)3 -> 2FeSO4 + CuSO4
b--------b
nFe2(SO4)3=0,2.1,1=0,22
=> a+b=0,22 (2)
Từ 1 và 2 => a=0,09 b=0,13
a/%mFe=0,09.56.100/15,28=32,98
%mCu=100-32,98=67,02
Bài 1 :
Đặt :
nCuO = x mol
nZnO = y mol
<=> 80x + 81y = 12.1 (1)
nHCl = 0.3 mol
CuO + 2HCl --> CuCl2 + H2O
x______2x
ZnO + 2HCl --> ZnCl2 + H2O
y_______2y
<=> x + y = 0.15 (2)
(1) , (2) :
x = 0.05
y = 0.1
mCuO = 4 g
mZnO = 8.1 g
%CuO = 33.05%
%ZnO = 66.95%
ZnO + H2SO4 --> ZnSO4 + H2O
0.1_____0.1
CuO + H2SO4 --> CuSO4 + H2O
0.05____0.05
mH2SO4 = 14.7 g
nCuCl2 : nZnCl2 = 0.05 : 0.1 = 1 : 2
Bài 2 :
Đặt :
nH2 = 0.2 mol
nAl = x mol
nFe = y mol
mAl + mFe = 7 - 1.5 = 5.5 g
<=> 27x + 56y = 5.5 (1)
2Al + 3H2SO4 --> Al2(SO4)3 + 3H2
x__________________________1.5x
Fe + H2SO4 --> FeSO4 + H2
y_____________________y
<=> 1.5x + y = 0.2 (2)
(1) , (2) :
x = 0.1
y = 0.05
mAl = 2.7 g
mFe = 2.8 g
mCu = 1.5 g
%Al = 38.57%
%Fe = 40%
%Cu = 21.43%
\(n_{CuO}=x;n_{ZnO}=y\)
\(PTHH:CuO+2HCl\rightarrow CuCl_2+H_2O\\ PTHH:ZnO+2HCl\rightarrow ZnCl_2+H_2O\)
\(\Rightarrow hpt:\left\{{}\begin{matrix}80x+81y=12,1\\2\left(x+y\right)=0,3\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=0,05\\y=0,1\end{matrix}\right.\)
\(\Rightarrow\left\{{}\begin{matrix}\%m_{CuO}=\frac{0,05.80}{12,1}.100\%=33,1\left(\%\right)\\\%m_{ZnO}=100-33,1=66,9\left(\%\right)\end{matrix}\right.\)
\(PTHH:CuO+H_2SO_4\rightarrow CuSO_4+H_2O\\ PTHH:ZnO+H_2SO_4\rightarrow ZnSO_4+H_2O\)
\(m_{H_2SO_4}=98.\left(0,05+0,1\right)=14,7\left(g\right)\)
\(\frac{n_{CuCl_2}}{n_{ZnCl_2}}=\frac{0,05}{0,1}=\frac{1}{2}\)