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\(n_{H_2}=\dfrac{8,96}{22,4}=0,4\left(mol\right)\\ PTHH:2Al+6HCl\rightarrow2AlCl_3+3H_2\\ Fe+2HCl\rightarrow FeCl_2+3H_2\\ Đặt:n_{Al}=a\left(mol\right);n_{Fe}=b\left(mol\right)\left(a,b>0\right)\\ \Rightarrow\left\{{}\begin{matrix}27a+56b=11\\1,5a+b=0,4\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}a=0,2\\b=0,1\end{matrix}\right.\\ a,n_{HCl}=2.n_{H_2}=2.0,4=0,8\left(mol\right)\\ \Rightarrow V_{ddHCl}=\dfrac{0,8}{8}=0,1\left(l\right)\\ b,FeCl_2+2AgNO_3\rightarrow Fe\left(NO_3\right)_2+2AgCl\downarrow\\ AlCl_3+3AgNO_3\rightarrow Al\left(NO_3\right)_3+3AgCl\downarrow\\ n_{AgCl}=n_{AgNO_3}=3.n_{AlCl_3}+2.n_{FeCl_2}=3.a+2.b=3.0,2+2.0,1=0,8\left(mol\right)\\ \Rightarrow a=\dfrac{170.0,8}{250}.100=54,4\%\\ b=m_{\downarrow}=m_{AgCl}=0,8.143,5=114,8\left(g\right)\)
\(n_{H2}=\dfrac{4,48}{22,4}=0,2\left(mol\right)\)
a) Pt : \(Zn+2HCl\rightarrow ZnCl_2+H_2|\)
1 2 1 1
0,2 0,2
b) \(n_{Zn}=\dfrac{0,2.1}{1}=0,2\left(mol\right)\)
\(m_{Zn}=0,2.65=13\left(g\right)\)
\(m_{Cu}=19,4-13=6,4\left(g\right)\)
Chúc bạn học tốt
a) Sửa đề: dd H2SO4 9,8%
Ta có: \(n_{H_2}=\dfrac{7,84}{22,4}=0,35\left(mol\right)\) \(\Rightarrow m_{H_2}=0,35\cdot2=0,7\left(g\right)\)
Bảo toàn nguyên tố: \(n_{H_2SO_4}=n_{H_2}=0,35\left(mol\right)\) \(\Rightarrow m_{ddH_2SO_4}=\dfrac{0,35\cdot98}{9,8\%}=350\left(g\right)\)
\(\Rightarrow m_{dd}=m_{KL}+m_{H_2SO_4}-m_{H_2}=361,6\left(g\right)\)
b) Tương tự câu a
a) PTHH: \(Zn+2HCl\rightarrow ZnCl_2+H_2\uparrow\)
b) Ta có: \(n_{H_2}=\dfrac{1,12}{22,4}=0,05\left(mol\right)\)
\(\Rightarrow n_{HCl}=0,1mol\) \(\Rightarrow C_{M_{HCl}}=\dfrac{0,1}{0,2}=0,5\left(M\right)\)
c) Theo PTHH: \(n_{Zn}=n_{H_2}=0,05mol\)
\(\Rightarrow m_{Zn}=0,05\cdot65=3,25\left(g\right)\)
\(\Rightarrow\%m_{Zn}=\dfrac{3,25}{8,37}\cdot100\%\approx38,83\%\) \(\Rightarrow\%m_{Cu}=61,17\%\)
\(n_{Al}=a;n_{Al_2O_3}=b\\ 2Al+6HCl\rightarrow2AlCl_3+3H_2\\ Al_2O_3+6HCl\rightarrow2AlCl_3+3H_2O\\ \Rightarrow\left\{{}\begin{matrix}27a+102b=23,1\\(a+2b)133,5=66,75\end{matrix}\right.\\ \Rightarrow a=0,1;b=0,2\\ \%m_{Al}=\dfrac{0,1.27}{23,1}\cdot100=11,7\%\\ \%m_{Al_2O_3}=100-11,7=88,3\%\)
Sửa đề: 3,785 (l) → 3,7185 (l)
a, \(2Al+6HCl\rightarrow2AlCl_3+3H_2\)
\(Al_2O_3+6HCl\rightarrow2AlCl_3+3H_2O\)
b, Ta có: \(n_{H_2}=\dfrac{3,7185}{24,79}=0,15\left(mol\right)\)
Theo PT: \(n_{Al}=\dfrac{2}{3}n_{H_2}=0,1\left(mol\right)\)
\(\Rightarrow\left\{{}\begin{matrix}\%m_{Al}=\dfrac{0,1.27}{40}.100\%=6,75\%\\\%m_{Al_2O_3}=93,25\%\end{matrix}\right.\)
c, \(n_{Al_2O_3}=\dfrac{40.93,25\%}{102}=\dfrac{373}{1020}\left(mol\right)\)
Theo PT: \(n_{HCl}=3n_{Al}+6n_{Al_2O_3}=\dfrac{212}{85}\left(mol\right)\)
\(\Rightarrow V_{HCl}=\dfrac{\dfrac{212}{85}}{2}=\dfrac{106}{85}\left(l\right)\approx1247,06\left(ml\right)\)
d, \(n_{AlCl_3}=n_{Al}+2n_{Al_2O_3}=\dfrac{212}{255}\left(mol\right)\)
\(\Rightarrow m_{AlCl_3}=\dfrac{212}{255}.133,5=\dfrac{9434}{85}\left(g\right)\)
e, \(C_{M_{AlCl_3}}=\dfrac{\dfrac{212}{255}}{\dfrac{106}{85}}=\dfrac{2}{3}\left(M\right)\)
a) \(n_{CO_2}=\dfrac{0,4958}{24,79}=0,02\left(mol\right);n_{HCl}=0,6.1=0,6\left(mol\right)\)
PTHH: \(CaCO_3+2HCl\rightarrow CaCl_2+CO_2+H_2O\)
0,02<------0,04<----0,02<-----0,02
\(\Rightarrow n_{HCl\left(p\text{ư}\right)}< n_{HCl\left(b\text{đ}\right)}\left(0,04< 0,6\right)\Rightarrow HCl\) dư, \(CaCO_3\) tan hết
\(\Rightarrow\left\{{}\begin{matrix}m_{CaCO_3}=0,02.100=2\left(g\right)\\m_{CaSO_4}=5-2=3\left(g\right)\end{matrix}\right.\)
b) dd sau phản ứng có: \(\left\{{}\begin{matrix}n_{HCl\left(d\text{ư}\right)}=0,6-0,04=0,56\left(mol\right)\\n_{CaCl_2}=0,02\left(mol\right)\end{matrix}\right.\)
\(\Rightarrow\left\{{}\begin{matrix}C_{M\left(HCl\left(d\text{ư}\right)\right)}=\dfrac{0,56}{0,6}=\dfrac{14}{15}M\\C_{M\left(CaCl_2\right)}=\dfrac{0,02}{0,6}=\dfrac{1}{30}M\end{matrix}\right.\)
0,3mol chứ nhỉ?
a. \(2Al+6HCl\rightarrow2AlCl_3+3H_2\uparrow\left(1\right)\)
\(FeO+2HCl\rightarrow FeCl_2+H_2O\left(2\right)\)
\(Fe_2O_3+6HCl\rightarrow2FeCl_3+3H_2O\left(3\right)\)
b. Theo phương trình \(n_{Al}=\frac{2}{3}n_{H_2}=0,2mol\) và \(n_{HCl\left(1\right)}=0,6mol\)
\(\rightarrow m_{FeO}+m_{Fe_2O_3}=35,8-0,2.27=30,4g\)
Đặt \(\hept{\begin{cases}n_{FeO}=x\\n_{Fe_2O_3}=y\end{cases}}\)
\(\rightarrow72x+160y=30,4\left(1\right)\)
Theo phương trình \(2x+6y=n_{HCl\left(2+3\right)}=1,6.1-0,6=1\left(2\right)\)
Từ (1) và (2) suy ra x = 0,2 và y = 0,1
\(\rightarrow m_{FeO}=0,2.72=14,4g\) và \(m_{Fe_2O_3}=0,1.160=16g\)
\(\rightarrow\%m_{FeO}=\frac{14,4}{35,8}.100\%\approx40,22\%\)
\(\rightarrow\%m_{Fe_2O_3}=\frac{16}{35,8}.100\%\approx44,69\%\)
c. Theo phương trình \(n_{AlCl_3}=0,2mol\) và \(n_{FeCl_2}=0,2mol\) và \(n_{FeCl_3}=0,2mol\)
\(\rightarrow m_{\text{muối}}=0,2.133,5+0,2.127+0,2.162,5=84,6g\)