Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
Gọi \(\left\{{}\begin{matrix}n_{CuO}=x\\n_{MgO}=y\end{matrix}\right.\)
\(CuO+2HCl\rightarrow CuCl_2+H_2O\)
x x ( mol )
\(MgO+2HCl\rightarrow MgCl_2+H_2O\)
y y ( mol )
Ta có:
\(\left\{{}\begin{matrix}80x+40y=16\\135x+95y=32,5\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}x=0,1\\y=0,2\end{matrix}\right.\)
\(\Rightarrow m_{CuO}=0,1.80=8g\)
\(\Rightarrow m_{MgO}=0,2.40=8g\)
\(\%m_{CuO}=\dfrac{8}{16}.100=50\%\)
\(\%m_{MgO}=\dfrac{8}{16}.100=50\%\)
\(m_{CuCl_2}=0,1.135=13,5g\)
\(m_{MgCl_2}=0,2.95=19g\)
nHCl = 0,3.0,3 = 0,09 (mol)
\(n_{H_2}=\dfrac{0,672}{22,4}=0,03\left(mol\right)\)
PTHH: 2Al + 6HCl --> 2AlCl3 + 3H2
0,02<-0,06<------------0,03
CuO + 2HCl --> CuCl2 + H2O
0,015<-0,03
=> \(\left\{{}\begin{matrix}m_{Al}=0,02.27=0,54\left(mol\right)\\m_{CuO}=0,015.80=1,2\left(g\right)\end{matrix}\right.\)
\(n_{HCl}=0,3\cdot0,3=0,09mol\)
\(n_{H_2}=\dfrac{0,672}{22,4}=0,03mol\)
\(CuO+2HCl\rightarrow CuCl_2+H_2O\)
\(2Al+6HCl\rightarrow2AlCl_3+3H_2\)
0,02 0,06 0,03
\(\Rightarrow n_{HCl\left(CuO\right)}=0,09-0,06=0,03mol\)
\(\Rightarrow n_{CuO}=n_{HCl}=0,03mol\) (theo pt)
\(\Rightarrow m_{CuO}=0,03\cdot80=2,4g\)
\(m_{Al}=0,02\cdot27=0,54g\)
\(n_{H_2}=\dfrac{0,672}{22,4}=0,03\left(mol\right)\)
PTHH:
2Al + 6HCl ---> 2AlCl3 + 3H2
0,02 0,06 0,03
nHCl = 0,3.0,3 = 0,09 (mol)
nHCl (CuO) = 0,09 - 0,06 = 0,03 (mol)
CuO + 2HCl ---> CuCl2 + H2O
0,015 0,03
\(\rightarrow\left\{{}\begin{matrix}m_{Al}=0,02.27=0,54\left(g\right)\\m_{CuO}=0,015.80=1,2\left(g\right)\end{matrix}\right.\)
P/s: mình có thấy chị Hương Giang làm nhưng sai phần tính số mol của CuO "\(n_{CuO}=n_{HCl}\) (theo pt)"
-
Fe2O3 + 6HCl --> 2FeCl3 + 3H2O
Mg + 2HCl --> MgCl2 + H2
-
\(n_{H_2}=\dfrac{4,48}{22,4}=0,2\left(mol\right)\)
\(n_{HCl}=\dfrac{200.18,25}{100.36,5}=1\left(mol\right)\)
PTHH: Mg + 2HCl --> MgCl2 + H2
0,2<----0,4<---------------0,2
Fe2O3 + 6HCl --> 2FeCl3 + 3H2O
0,1<-----0,6
=> \(\left\{{}\begin{matrix}\%Mg=\dfrac{0,2.24}{0,2.24+0,1.160}.100\%=23,077\%\\\%Fe_2O_3=\dfrac{0,1.160}{0,2.24+0,1.160}.100\%=76,923\%\end{matrix}\right.\)
PTHH: \(Zn+2HCl\rightarrow ZnCl_2+H_2\uparrow\)
\(Al_2O_3+6HCl\rightarrow2AlCl_3+3H_2O\)
a) Ta có: \(n_{H_2}=\dfrac{8,96}{22,4}=0,4\left(mol\right)=n_{Zn}\)
\(\Rightarrow\%m_{Zn}=\dfrac{0,4\cdot65}{36,2}\cdot100\%\approx71,23\%\) \(\Rightarrow\%m_{Al_2O_3}=28,77\%\)
c) Ta có: \(n_{Al_2O_3}=\dfrac{36,2-0,4\cdot65}{102}=0,1\left(mol\right)\)
Theo PTHH: \(n_{HCl}=2n_{Zn}+6n_{Al_2O_3}=1,4\left(mol\right)\)
\(\Rightarrow m_{ddHCl}=\dfrac{1,4\cdot36,5}{10\%}=511\left(g\right)\) \(\Rightarrow V_{ddHCl}=\dfrac{511}{1,1}\approx464,5\left(ml\right)=0,4645\left(l\right)\)
c) Theo PTHH: \(\left\{{}\begin{matrix}n_{ZnCl_2}=0,4\left(mol\right)\\n_{AlCl_3}=0,2\left(mol\right)\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}C_{M_{ZnCl_2}}=\dfrac{0,4}{0,4645}\approx0,86\left(M\right)\\C_{M_{AlCl_3}}=\dfrac{0,2}{0,4645}\approx0,43\left(M\right)\end{matrix}\right.\)
nHCl=0,4.0,1=0,04(mol)
Đặt : nCaO=x(mol); nCuO=y(mol)
\(\Rightarrow\) 56x+80y=1,36(1)
PT: CaO+2HCl \(\rightarrow\) CaCl2+H2O
x 2x x (mol)
CuO+2HCl \(\rightarrow\) CuCl2+H2O
y 2y y (mol)
nHCl=0,04 mol \(\Rightarrow\) 2x+2y=0,04(2)
Từ (1) và (2) giải hệ pt tìm đk : x=y=0,01
%m CaO=41,2% ; %mCuO=58,8%
mCaCl2=0,01.111=1,11g
mCuCl2=0,01.135=1,35 g
a. PTHH: CaO + 2HCL suy ra CaCL2 +H20 (1)
x 2x x (mol)
CuO + 2HCL suy ra CuCL2 +H2O (2)
y 2y y (mol)
Gọi nCaO=x (mol) theo pt(1): nHCL=2x (mol)
nCuO=y (mol) theo pt(2): nHCL=2y (mol)
Theo pt (1) và (2) ta có :
mCaO+mCuO =mhh hay 56x + 80y =1.36 (g) (*)
Mặt khác ta lại có : nHCL = 2x+2y=0.04 (mol) (**)
Gỉai (*) và (**) ta đc : x=0.01(mol) ; y=0.01(mol)
nCaO=x=0.01(mol) mCaO=0.01 .56=0.56(g)
nCuO=y=0.01(mol)mCuO=0.01 . 80 =0.8(g)
%CaO=
41.18%
%CuO=
58.82%
b., Theo pthh (1) và (2) :nCaO=nCaCL2=x=0.01(mol)
nCuO=nCuCL2=y=0.01(mol)
mCaCL2=0.01.111=1.11(g)
mCuCL2=0.01 . 135=1.35 (g)