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Gọi \(\left\{{}\begin{matrix}n_{CuO}=x\\n_{Fe_2O_3}=y\end{matrix}\right.\)
\(n_{HCl}=0,5.2=1mol\)
\(CuO+2HCl\rightarrow CuCl_2+H_2\)
x 2x ( mol )
\(Fe_2O_3+6HCl\rightarrow2FeCl_3+3H_2O\)
y 6y ( mol )
Ta có:
\(\left\{{}\begin{matrix}80x+160y=32\\2x+6y=1\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}x=0,2\\y=0,1\end{matrix}\right.\)
\(\Rightarrow m_{CuO}=0,2.80=16g\)
\(\Rightarrow m_{Fe_2O_3}=0,1.160=16g\)
Gọi \(\left\{{}\begin{matrix}n_{CuO}=a\left(mol\right)\\n_{Fe_2O_3}=b\left(mol\right)\end{matrix}\right.\)
PTHH:
CuO + HCl ---> CuCl2 + H2O
a 2a
Fe2O3 + 6HCl ---> 2FeCl3 + 3H2O
b 6b
\(n_{HCl}=2.0,5=1\left(mol\right)\)
Hệ phương trình \(\left\{{}\begin{matrix}80a+160b=32\\2a+6b=1\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}a=0,2\left(mol\right)\\b=0,1\left(mol\right)\end{matrix}\right.\)
\(\rightarrow\left\{{}\begin{matrix}m_{CuO}=0,2.80=16\left(g\right)\\m_{Fe_2O_3}=0,1.160=16\left(g\right)\end{matrix}\right.\)
Bài 1:
\(2Al+6HCl\rightarrow2AlCl_3+3H_2\\ n_{H_2}=\dfrac{6,72}{22,4}=0,3\left(mol\right)\\ Al_2O_3+6HCl\rightarrow2AlCl_3+3H_2\\ n_{Al}=\dfrac{2}{3}.0,3=0,2\left(mol\right)\\ \Rightarrow m_{Al}=0,2.27=5,4\left(g\right)\\ \%m_{Al}=\dfrac{5,4}{26,82}.100\approx20,134\%\\\Rightarrow \%m_{Al_2O_3}\approx79,866\%\\ b,n_{Al_2O_3}=\dfrac{26,82-5,4}{102}=0,21\left(mol\right)\\ n_{HCl}=6.0,21+2.0,3=1,86\left(mol\right)\\ V_{ddHCl}=\dfrac{1,86}{2}=0,93\left(l\right)=930\left(ml\right)\\ m_{ddHCl}=930.1,12=1041,6\left(g\right)\\ n_{AlCl_3}=2.0,21+0,2=0,62\left(mol\right)\\ C\%_{ddAlCl_3}=\dfrac{0,62.133,5}{1041,6-0,3.2}.100\approx7,951\%\)
2)
a) Gọi KL và oxit của nó là M và MO
nHCl = 4.0,25 = 1 (mol)
\(n_{H_2}=\dfrac{6,72}{22,4}=0,3\left(mol\right)\)
PTHH: M + 2HCl --> MCl2 + H2
0,3<-0,6<--------------0,3
MO + 2HCl --> MCl2 + H2O
0,2<---0,4
=> 0,3.MM + 0,2.(MM + 16) = 31,2
=> MM = 56 (g/mol)
=> Kim loại là Sắt (Fe)
b)
\(\left\{{}\begin{matrix}\%m_{Fe}=\dfrac{0,3.56}{31,2}.100\%=53,85\%\\\%m_{FeO}=\dfrac{0,2.72}{31,2}.100\%=46,15\%\end{matrix}\right.\)
\(m_O=\dfrac{15.12,8}{100}=1,92\left(g\right)\)
=> \(n_O=\dfrac{1,92}{16}=0,12\left(mol\right)\)
=> \(n_{H_2O}=0,12\left(mol\right)\)
\(n_{H_2}=\dfrac{3,36}{22,4}=0,15\left(mol\right)\)
Bảo toàn H: nHCl = 0,12.2 + 0,15.2 = 0,54 (mol)
=> nCl = 0,54 (mol)
mmuối = mhh rắn - mO + mCl
= 15 - 1,92 + 0,54.35,5 = 32,25 (g)
a) 2Al + 6HCl -> 2AlCl3 + 3H2
Al2O3 + 6HCl -> 2AlCl3 + 3H2O
nH2 = 0,15mol => nAl=0,1mol => mAl=2,7g; mAl2O3 = 10,2g => nAl2O3 = 0,1mol
=>%mAl=20,93% =>%mAl2O3 = 79,07%
b) nHCl = 0,1.3+0,1.6=0,9 mol=>mHCl(dd)=100g
mddY=12,9+100-0,15.2=112,6g
mAlCl3=22,5g=>C%=19,98%
Gọi \(\left\{{}\begin{matrix}n_{H_2}=a\left(mol\right)\\n_{CO_2}=b\left(mol\right)\end{matrix}\right.\)
=> \(\left\{{}\begin{matrix}a+b=\dfrac{4,704}{22,4}=0,21\\\overline{M}=\dfrac{2a+44b}{a+b}=12,5.2=25\end{matrix}\right.\)
=> a = 0,095 (mol); b = 0,115 (mol)
Đặt nHCl = x (mol)
\(n_{MgCl_2}=\dfrac{12,825}{95}=0,135\left(mol\right)\)
Bảo toàn Cl: \(n_{CaCl_2}=\dfrac{x-0,27}{2}\left(mol\right)\)
Bảo toàn H: \(n_{H_2O}=\dfrac{x-2.0,095}{2}=\dfrac{x-0,19}{2}\left(mol\right)\)
BTKL:
\(m_{hh\left(bđ\right)}+m_{HCl}=m_{MgCl_2}+m_{CaCl_2}+m_{H_2}+m_{CO_2}+m_{H_2O}\)
=> \(19,02+36,5x=12,825+\dfrac{x-0,27}{2}.111+0,095.2+0,115.44+\dfrac{x-0,19}{2}.18\)
=> x = 0,63 (mol)
=> \(n_{CaCl_2}=0,18\left(mol\right)\)
=> mCaCl2 = 0,18.111 = 19,98 (g)
Đặt \(\begin{cases} n_{MgO}=x(mol)\\ n_{Al_2O_3}=y(mol) \end{cases} \Rightarrow 40x+102y=16,2(1)\)
\(n_{HCl}=0,45.2=0,9(mol)\\ PTHH:MgO+2HCl\to MgCl_2+H_2O\\ Al_2O_3+6HCl\to 2AlCl_3+3H_2O\\ \Rightarrow 2x+6y=0,9(2)\\ (1)(2)\Rightarrow \begin{cases} x=0,15(mol)\\ y=0,1(mol) \end{cases}\Rightarrow \begin{cases} m_{MgO}=40.0,15=6(g)\\ m_{Al_2O_3}=102.0,1=10,2(g) \end{cases} \)