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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%
\(n_{HCl} = \dfrac{448.1,12.3,65\%}{36,5} = 0,50176(mol)\\ Zn + 2HCl \to ZnCl_2 + H_2\\ ZnO + 2HCl \to ZnCl_2 + H_2O\\ n_{Zn} = n_{H_2} = \dfrac{2,24}{22,4} = 0,1(mol)\\ n_{ZnO} = \dfrac{n_{HCl} - 2n_{Zn}}{2} = \dfrac{0,50176-0,1.2}{2} = 0,15088(mol)\\ \%m_{Zn} = \dfrac{0,1.65}{0,1.65 + 0,15088.81}.100\% = 34,72\%\\ \%m_{ZnO} = 65,28\%\)
\(Zn+2HCl\rightarrow ZnCl_2+H_2\)
0,2___0,4_____ 0,2_____ 0,2
\(ZnO+2HCl\rightarrow ZnCl_2+H_2O\)
0,1____0,2 _____0,1
\(n_{H2}=\frac{4,48}{22,4}=0,2\left(mol\right)\)
\(m_{ZnO}=21,1-0,2.65=8,1\left(g\right)\)
\(\rightarrow n_{ZnO}=\frac{8,1}{65+16}=0,1\left(mol\right)\)
Đổi : 600ml==0,6lit
\(CM_{HCl}=\frac{0,6}{0,6}=1\left(M\right)\)
\(\rightarrow m_{ZnCl2}=0,3\left(65+71\right)=40,8\left(g\right)\)
Zn + HCl = ZnCl2 + H2 (1)
ZnO + HCl = ZnCl2 + H2O (2)
a,
nH2= nZn = 0.2 (mol) => mZn = 0.2*65=13(g)
=> mZnO= 21,1 - 13 = 8.1 (g)
%mZn= 13/21,1*100 = 61,6% => %mZnO = 38,4%
b,
nHCl (1)= 2nH2=0,4 (mol)
nHCl(2) = 2nZnO = 2*(8,1/81) = 0,2 (mol)
=>nHCl = nHCl(1) + nHCl(2) = 0,6 (mol)
CMHCl = \(\frac{n}{V}\)= 1 M
c,
nCl = nHCl = 0,6 (mol)
mmuối = mhh + mCl = 21.1 + ( 0.6*35,5) =42,4 (g)
\(a) n_{Mg}= a(mol) ; n_{Al} = b(mol) \Rightarrow 24a + 27b =2,55(1)\\ Mg + 2HCl \to MgCl_2 + H_2\\ 2Al + 6HCl \to 2AlCl_3 + 3H_2\\ n_{H_2} = a + 1,5b = \dfrac{2,8}{22,4}=0,125(2)\\ (1)(2) \Rightarrow a = b = 0,05\\ \%m_{Mg} = \dfrac{0,05.24}{2,55}.100\% = 47,06\%\ ;\ \%m_{Al} =100\% -47,06\% = 52,94\%\\ b) n_{HCl} = 2n_{H_2} = 0,125.2 = 0,25(mol)\\ m_{dd\ HCl} = \dfrac{0,25.36,5}{7,3\%} = 125(gam)\\ V_{dd\ HCl} = \dfrac{125}{1,2} = 104,17(ml)\)
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: \(\left\{{}\begin{matrix}n_{ZnCl_2}=0,4\left(mol\right)\\n_{AlCl_3}=0,2\left(mol\right)\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}C_{M_{ZnCl_2}}=\dfrac{0,4}{0,4645}\approx0,86\left(M\right)\\C_{M_{AlCl_3}}=\dfrac{0,2}{0,4645}\approx0,43\left(M\right)\end{matrix}\right.\)
1, Zn + 2HCl--> ZnCl2 + H2(1)
ZnO + 2HCl--> ZnCl2 + H2O(2)
Ta có m dd HCl=D.V=100,8.1,19=119,952g
=> mHCl=36,5.119,952/100=43,78248 g
=> nHCl=43,78248/36,5=1,19952 mol
Ta có : nZn=nH2=0,4mol
nHCl(1)=2nH2=0,8mol
=> nHCl(2)=1,19952-0,8=0,39952mol
ta có nHCl(2)/2=nZnO=0,19976mol
=> %mZn=0,4.65.100/(0,4.65+0,19976.81)=67,39%
=> %mZnO=100-67,39=32,61%
Chúc bạn hk tốt
đề có cho bt h2 vs h2o
bao nhiu mol đâu mà sao bn biết nHCL (1 ) = 2nH2 = 0,8 thế ???
a, Ta có: 24nMg + 56nFe = 9,2 (g) (1)
\(n_{H_2}=\dfrac{5,6}{22,4}=0,25\left(mol\right)\)
BT e, có: 2nMg + 2nFe = 2nH2 = 0,5 (2)
Từ (1) và (2) \(\Rightarrow\left\{{}\begin{matrix}n_{Mg}=0,15\left(mol\right)\\n_{Fe}=0,1\left(mol\right)\end{matrix}\right.\)
\(\Rightarrow\left\{{}\begin{matrix}\%m_{Mg}=\dfrac{0,15.24}{9,2}.100\%\approx39,13\%\\\%m_{Fe}\approx60,87\%\end{matrix}\right.\)
b, BTNT H, có: \(n_{HCl}=2n_{H_2}=0,5\left(mol\right)\Rightarrow C_{M_{HCl}}=\dfrac{0,5}{0,2}=2,5\left(M\right)\)
Đáp án A
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)
%mZnO = 100% -61,6% = 38,4%