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\(a,PTHH:Zn+2HCl\to ZnCl_2+H_2\\ ZnO+2HCl\to ZnCl_2+H_2O\\ b,n_{H_2}=\dfrac{3,36}{22,4}=0,15(mol)\\ \Rightarrow n_{Zn}=0,15(mol)\Rightarrow m_{Zn}=0,15.65=9,75(g)\\ \Rightarrow \%_{Zn}=\dfrac{9,75}{25,95}.100\%=37,57\%\\ \Rightarrow \%_{ZnO}=(100-37,57)\%=62,43\%\\ c,n_{ZnO}=\dfrac{25,95-9,75}{81}=0,2(mol)\\ \Rightarrow n_{HCl}=2.0,15+2.0,2=0,7(mol)\\ \Rightarrow m_{dd_{HCl}}=\dfrac{0,7.36,5}{12\%}=212,92(g)\)
\(n_{H_2}=\dfrac{2,24}{22,4}=0,1\left(mol\right)\)
a.
\(CuO+2HCl\rightarrow CuCl_2+H_2O\)
\(Mg+2HCl\rightarrow MgCl_2+H_2\)
b.
\(n_{Mg}=n_{H_2}=0,1\left(mol\right)\Rightarrow m_{Mg}=0,1.24=2,4\left(g\right),m_{CuO}=12-2,4=9,6\left(g\right)\)
c.
\(m_{muối}=m_{CuCl_2}+m_{MgCl_2}=0,12.135+95.0,1=25,7\left(g\right)\)
a) Pt : \(Mg+2HCl\rightarrow MgCl_2+H_2|\)
1 2 1 1
0,3 0,6 0,3
\(MgO+2HCl\rightarrow MgCl_2+H_2O|\)
1 2 1 1
0,1 0,2
b) \(n_{Mg}=\dfrac{0,3.1}{1}=0,3\left(mol\right)\)
\(m_{Mg}=0,3.24=7,2\left(g\right)\)
\(m_{MgO}=11,2-7,2=4\left(g\right)\)
c) 0/0Mg = \(\dfrac{7,2.100}{11,2}=64,29\)0/0
0/0MgO = \(\dfrac{4.100}{11,2}=35,71\)0/0
d) Có : \(m_{MgO}=4\left(g\right)\)
\(n_{MgO}=\dfrac{4}{40}=0,1\left(mol\right)\)
\(n_{HCl\left(tổng\right)}=0,6+0,2=0,8\left(mol\right)\)
\(m_{HCl}=0,8.36,5=29,2\left(g\right)\)
\(C_{ddHCl}=\dfrac{29,2.100}{200}=14,6\)0/0
Chúc bạn học tốt
\(a/ZnO+2HCl\rightarrow ZnCl_2+H_2O\\ Fe_2O_3+6HCl\rightarrow2FeCl_3+3H_2O\\ b/n_{HCl}=\dfrac{250.7,3}{100}:36,5=0,5mol\\ n_{ZnO}=a;n_{Fe_2O_3}=b\\ \Rightarrow\left\{{}\begin{matrix}81a+160b=16,1\\2a+6b=0,5\end{matrix}\right.\\ \Rightarrow a=b=0,1mol\\ \%m_{ZnO}=\dfrac{0,1.81}{16,1}\cdot100=50,3\%\\ \%m_{Fe_2O_3}=100-50,3=49,7\%\\ c/C_{\%ZnCl_2}=\dfrac{136.0,1}{16,1+250}\cdot100=5\%\\ C_{\%FeCl_3}=\dfrac{0,1.2.162,5}{16,1+250}=12\%\)
a) PTHH: Fe + 2HCl ===> FeCl2 + H2
MgO + 2HCl ===> MgCl2 + H2O
b) Ta có: nH2 =\(\frac{2,24}{22,4}=0,1\left(mol\right)\)
=> nFe = nH2 = 0,1 (mol)
=> mFe = 0,1 x 56 = 5,6 gam
=> mMgO = 22,4 - 5,6 = 156,8 gam
c. %MgO = \(\frac{16,8}{22,4}.100\%=75\%\)
=> %Fe = 100% - 75% = 25%
nH2=0,1(mol)
PTHH: Mg + 2 HCl -> MgCl2 + H2
0,1__________0,2___________0,1(mol)
MgO + 2 HCl -> MgCl2 + H2O
0,05____0,1___0,05(mol)
mMg=0,1. 24= 2,4(g) -> mMgO=4,4-2,4= 2(g) -> nMgO=0,05((mol)
b) %mMg= (2,4/4,4).100=54,545%
=> %mMgO=45,455%
c) nHCl=0,3(mol) -> mHCl=0,3.36,5=10,95(g)
=> mddHCl=(10,95.100)/7,3=150(g)
Lượng khí hidro thoát ra là 0,15 mol.
a) Al2O3 + 2NaOH \(\rightarrow\) 2NaAlO2 + H2O
2Al + 2NaOH + 2H2O \(\rightarrow\) 2NaAlO2 + 3H2.
b) Lượng Al phản ứng bằng 2/3 số mol khí hidro thoát ra và bằng 0,1 mol.
\(\Rightarrow\) %mnhôm=\(\dfrac{27.0,1}{12,9}\).100%\(\approx\)20,93%, %mnhôm oxit\(\approx\)100%-20,93%\(\approx\)79,07%.
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)"
a,PTHH:
\(Fe_2O_3+6HCl\rightarrow2FeCl_3+3H_2O\)
x 6x 2x 3x
\(ZnO+2HCl\rightarrow ZnCl_2+H_2O\)
y 2y y y
\(\left\{{}\begin{matrix}160x+81y=16,1\\325x+136y=29,85\end{matrix}\right.\)
\(\Leftrightarrow x=0,05;y=0,1\)
b,\(\%m_{Fe_2O_3}=\dfrac{0,05.160}{16,1}.100\%=49,69\%\)
\(\%m_{ZnO}=100\%-49,69=50,31\%\)
c, \(C_{M\left(HCl\right)}=\dfrac{0,5}{0,3}=\dfrac{5}{3}\left(M\right)\)