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nH2 = 6,72/22,4 = 0,3 (mol)
PTHH: 2Al + 6HCl -> 2AlCl3 + 3H2
nAl = 0,3 : 3 . 2 = 0,2 (mol)
nHCl (Al) = 0,3 . 2 = 0,6 (mol)
mAl = 0,2 . 27 = 5,4 (g)
%mAl = 5,4/25,65 = 20,05%
%mZnO = 100% - 20,05% = 79,95%
mZnO = 25,65 - 5,4 = 20,25 (g)
nZnO = 20,25/81 = 0,25 (mol)
PTHH: ZnO + 2HCl -> ZnCl2 + H2O
nHCl (ZnO) = 0,25 . 2 = 0,5 (mol)
nHCl (đã dùng) = 0,6 + 0,5 = 1,1 (mol)
CMddHCl = 1,1/0,1008 = 10,9M
C% = (10,9 . 36,5)/(10 . 1,19) = 33,43%
\(mdd_{HCl}=100,8.1,19=119,952\left(g\right)\)
\(\Rightarrow m_{HCl}=\dfrac{119,952\times36}{100}=43,18272\left(g\right)\)
\(\Rightarrow n_{HCl}=\dfrac{43,18272}{36,5}=1,18\left(mol\right)\)
\(n_{H_2}=\dfrac{8,96}{22,4}=0,4\left(mol\right)\)
PTHH: \(Zn+2HCl\rightarrow ZnCl_2+H_2\left(1\right)\)
\(ZnO+2HCl\rightarrow ZnCl_2+H_2O\left(2\right)\)
Theo PT (1) ta có: \(n_{Zn}=n_{H_2}=0,4\left(mol\right)\)
\(\Rightarrow m_{Zn}=0,4.65=26\left(g\right)\)
Theo PT (1): \(n_{HCl\left(PT1\right)}=\dfrac{0,4.2}{1}=0,8\left(mol\right)\)
\(\Rightarrow n_{HCl\left(PT2\right)}=1,18-0,8=0,38\left(mol\right)\)
Theo PT (2): \(n_{ZnO}=\dfrac{0,38.1}{2}=0,19\left(mol\right)\)
\(\Rightarrow m_{ZnO}=0,19.81=15,39\left(g\right)\)
\(m_{hỗn-hợp-ban-đầu}=26+15,39=41,39\left(g\right)\)
\(\Rightarrow\%_{Zn}=\dfrac{26}{41,39}.100\%=62,81\%\)
\(\Rightarrow\%ZnO=100\%-62,81\%=37,19\%\)
Đáp án A
.100 => mHCl = 43,78 (g)
nHCl = 1,2 (mol)
Gọi nZn = a, nZnO = b
Zn + 2HCl → ZnCl2 + H2
0,4 0,8 ← 0,4 (mol)
ZnO + 2HCl → ZnCl2 + H2O
0,2 ← 0,4 (mol)
.100%
.100% = 61,61%
%mZnO = 100% -61,6% = 38,4%
\(n_{HCl} = \dfrac{448.1,12.3,65\%}{36,5} = 0,50176(mol)\\ Zn + 2HCl \to ZnCl_2 + H_2\\ ZnO + 2HCl \to ZnCl_2 + H_2O\\ n_{Zn} = n_{H_2} = \dfrac{2,24}{22,4} = 0,1(mol)\\ n_{ZnO} = \dfrac{n_{HCl} - 2n_{Zn}}{2} = \dfrac{0,50176-0,1.2}{2} = 0,15088(mol)\\ \%m_{Zn} = \dfrac{0,1.65}{0,1.65 + 0,15088.81}.100\% = 34,72\%\\ \%m_{ZnO} = 65,28\%\)
\(a) n_{Mg}= a(mol) ; n_{Al} = b(mol) \Rightarrow 24a + 27b =2,55(1)\\ Mg + 2HCl \to MgCl_2 + H_2\\ 2Al + 6HCl \to 2AlCl_3 + 3H_2\\ n_{H_2} = a + 1,5b = \dfrac{2,8}{22,4}=0,125(2)\\ (1)(2) \Rightarrow a = b = 0,05\\ \%m_{Mg} = \dfrac{0,05.24}{2,55}.100\% = 47,06\%\ ;\ \%m_{Al} =100\% -47,06\% = 52,94\%\\ b) n_{HCl} = 2n_{H_2} = 0,125.2 = 0,25(mol)\\ m_{dd\ HCl} = \dfrac{0,25.36,5}{7,3\%} = 125(gam)\\ V_{dd\ HCl} = \dfrac{125}{1,2} = 104,17(ml)\)
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: \(\left\{{}\begin{matrix}n_{ZnCl_2}=0,4\left(mol\right)\\n_{AlCl_3}=0,2\left(mol\right)\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}C_{M_{ZnCl_2}}=\dfrac{0,4}{0,4645}\approx0,86\left(M\right)\\C_{M_{AlCl_3}}=\dfrac{0,2}{0,4645}\approx0,43\left(M\right)\end{matrix}\right.\)
a) Đặt \(\left\{{}\begin{matrix}n_{Mg}=a\left(mol\right)\\n_{Al}=b\left(mol\right)\end{matrix}\right.\) \(\Rightarrow24a+27b=5,1\) (1)
Ta có: \(n_{H_2}=\dfrac{5,6}{22,4}=0,25\left(mol\right)\)
Bảo toàn electron: \(2a+3b=0,5\) (2)
Từ (1) và (2) \(\Rightarrow\left\{{}\begin{matrix}a=0,1\\b=0,1\end{matrix}\right.\)
\(\Rightarrow\left\{{}\begin{matrix}\%m_{Mg}=\dfrac{0,1\cdot24}{5,1}\cdot100\%\approx47,06\%\\\%m_{Al}=52,94\%\end{matrix}\right.\)
b) Bảo toàn nguyên tố: \(n_{HCl}=2n_{H_2}=0,5\left(mol\right)\)
\(\Rightarrow m_{ddHCl}=\dfrac{0,5\cdot36,5}{7,3\%}=250\left(g\right)\)
\(\Rightarrow V_{HCl}=\dfrac{250}{1,2}\approx208,33\left(ml\right)\)
Fe+2HCl->FeCl2+H2
x-------------------------x mol
Zn+2HCl->ZnCl2+H2
y-------------------------y mol
=>Ta có hệ :\(\left\{{}\begin{matrix}56x+65y=14,9\\x+y=0,25\end{matrix}\right.\)
=>\(\left\{{}\begin{matrix}x=0,15\\y=0,1\end{matrix}\right.\)
=>%m Fe=\(\dfrac{0,15.56}{14,9}.100=56,375\%\)
=>VHCl=\(\dfrac{0,15.2+0,1.2}{2}=0,25l=250ml\)
Zn+2HCl--->ZnCl2+H2
Zn+2HCl--->ZnCl2+H2
Ta có
m dd HCl=100,8.1,19=119,952(g)
m HCl=119,952.36/100=43,18(g)
n HCl=43,18/36,5=1,183(mol)
Mà n H2=8,96/22,4=0,4(mol)
Theo pthh1
n HCl=2n H2=0,8(mol)
m HCl ở Pt 2=1,183-0,8=0,383(mol)
Theo pthh2
n ZnO=1/2n HCl=0,1915(mol)
m ZnO=0,1915.81=15,5115(g)
m Zn=0,4.65=26(g)
m Zn+m ZnO=26+15,5115=41,5115(g)
%m ZnO=15,5115/41,5115.100%=37,37%
Chúc bạn học tốt
PTHH ( I ) : \(Zn+2HCl\rightarrow ZnCl_2+H_2\)
PTHH ( II ) : \(ZnO+2HCl\rightarrow ZnCl_2+H_2O\)
\(m_{ddHCl}=D_{HCl}.V_{HCl}=100,8.1,19=119,952\left(g\right)\)
=> \(m_{HCl}=\frac{C\%_{HCl}.m_{ddHCl}}{100\%}=\frac{36\%.119,952}{100\%}=43,18272\left(g\right)\)
=> \(n_{HCl}=\frac{m_{HCl}}{M_{HCl}}=\frac{43,18272}{1+35,5}\approx1,18\left(mol\right)\)
Mà \(n_{HCl}=n_{HCl\left(I\right)}+n_{HCl\left(II\right)}\)
=> \(2n_{Zn}+2n_{ZnO}=1,18\) ( I )
\(n_{H_2}=\frac{V_{H_2}}{22,4}=\frac{8,96}{22,4}=0,4\left(mol\right)\)
Theo PTHH ( I ) : \(n_{Zn}=n_{H2}=0,4\left(mol\right)\)
=> \(m_{Zn}=n.M=0,4.65=26\left(g\right)\)
Thay \(n_{Zn}=0,4\) vào phương trình ( I ) ta được :
\(2.0,4+2n_{ZnO}=1,18\)
=> \(n_{ZnO}=0,19\left(mol\right)\)
=> \(m_{ZnO}=n.M=0,19.\left(65+16\right)=15,39\left(g\right)\)
Ta có : \(m_{hh}=m_{Zn}+m_{ZnO}=26+15,39=41,39\left(g\right)\)
=> \(\%ZnO=\frac{15,39}{41,39}.100\%\approx37,18\%\)