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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%
\(n_{HCl} = \dfrac{448.1,12.3,65\%}{36,5} = 0,50176(mol)\\ Zn + 2HCl \to ZnCl_2 + H_2\\ ZnO + 2HCl \to ZnCl_2 + H_2O\\ n_{Zn} = n_{H_2} = \dfrac{2,24}{22,4} = 0,1(mol)\\ n_{ZnO} = \dfrac{n_{HCl} - 2n_{Zn}}{2} = \dfrac{0,50176-0,1.2}{2} = 0,15088(mol)\\ \%m_{Zn} = \dfrac{0,1.65}{0,1.65 + 0,15088.81}.100\% = 34,72\%\\ \%m_{ZnO} = 65,28\%\)
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%
a .
\(Zn+2HCl\rightarrow ZnCl_2+H_2\uparrow\left(1\right)\left(Pu-oxi-hoa\right)\)
\(ZnO+2HCl\rightarrow ZnCl_2+H_2\left(2\right)\)
b. \(n_{H2}=\frac{4,48}{22,4}=0,2\left(mol\right)\)
\(\rightarrow n_{HCl\cdot\left(1\right)}=2nH_2=0,4\left(mol\right)=n_{Zn}\rightarrow m_{Zn}=0,4.65=26\left(g\right)\)
Mà \(n_{HCl}=0,6.1=0,6\left(mol\right)\)
\(n_{HCl\left(2\right)}=0,6-0,4=0,2\left(mol\right)\)
\(\rightarrow n_{ZnO}=\frac{1}{2}n_{HCl}=0,1\left(mol\right)\rightarrow m_{ZnO}=0,1.81=8,1\left(g\right)\)
\(n_{H_2}=\dfrac{6,72}{22,4}=0,3mol\)
\(\left\{{}\begin{matrix}n_{Mg}=x\left(mol\right)\\n_{Zn}=y\left(mol\right)\end{matrix}\right.\Rightarrow24x+65y=11,3\left(1\right)\)
\(Mg+2HCl\rightarrow MgCl_2+H_2\)
\(Zn+2HCl\rightarrow ZnCl_2+H_2\)
\(\Rightarrow x+y=0,3\left(2\right)\)
Từ (1) và (2) \(\Rightarrow\left\{{}\begin{matrix}x=0,2mol\\y=0,1mol\end{matrix}\right.\)
a)\(\%m_{Mg}=\dfrac{0,2\cdot24}{11,3}\cdot100\%=42,48\%\)
\(\%m_{Zn}=100\%-42,48\%=57,52\%\)
b)\(n_{HCl}=2\left(n_{Mg}+n_{Zn}\right)=2\cdot\left(0,2+0,1\right)=0,6mol\)
\(C_{M_{HCl}}=\dfrac{0,6}{0,2}=3M\)
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: \(\left\{{}\begin{matrix}n_{ZnCl_2}=0,4\left(mol\right)\\n_{AlCl_3}=0,2\left(mol\right)\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}C_{M_{ZnCl_2}}=\dfrac{0,4}{0,4645}\approx0,86\left(M\right)\\C_{M_{AlCl_3}}=\dfrac{0,2}{0,4645}\approx0,43\left(M\right)\end{matrix}\right.\)
\(Zn+2HCl\rightarrow ZnCl_2+H_2\)
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)