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\(n_{H_2}=\dfrac{8,96}{22,4}=0,4\left(mol\right)\\ PTHH:2Al+6HCl\rightarrow2AlCl_3+3H_2\\ Fe+2HCl\rightarrow FeCl_2+3H_2\\ Đặt:n_{Al}=a\left(mol\right);n_{Fe}=b\left(mol\right)\left(a,b>0\right)\\ \Rightarrow\left\{{}\begin{matrix}27a+56b=11\\1,5a+b=0,4\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}a=0,2\\b=0,1\end{matrix}\right.\\ a,n_{HCl}=2.n_{H_2}=2.0,4=0,8\left(mol\right)\\ \Rightarrow V_{ddHCl}=\dfrac{0,8}{8}=0,1\left(l\right)\\ b,FeCl_2+2AgNO_3\rightarrow Fe\left(NO_3\right)_2+2AgCl\downarrow\\ AlCl_3+3AgNO_3\rightarrow Al\left(NO_3\right)_3+3AgCl\downarrow\\ n_{AgCl}=n_{AgNO_3}=3.n_{AlCl_3}+2.n_{FeCl_2}=3.a+2.b=3.0,2+2.0,1=0,8\left(mol\right)\\ \Rightarrow a=\dfrac{170.0,8}{250}.100=54,4\%\\ b=m_{\downarrow}=m_{AgCl}=0,8.143,5=114,8\left(g\right)\)
Câu 1:
Gọi : nMg=a(mol); nMgO=b(mol) (a,b>0)
a) PTHH: Mg + 2 HCl -> MgCl2 + H2
a________2a_______a______a(mol)
MgO +2 HCl -> MgCl2 + H2O
b_____2b_______b___b(mol)
Ta có hpt:
\(\left\{{}\begin{matrix}24a+40b=8,8\\22,4a=4,48\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}a=0,2\\b=0,1\end{matrix}\right.\)
=> mMg=0,2.24=4,8(g)
=>%mMg= (4,8/8,8).100=54,545%
=> %mMgO= 45,455%
b) m(muối)=mMg2+ + mCl- = 0,3. 24 + 0,6.35,5=28,5(g)
c) V=VddHCl=(2a+2b)/2=0,3(l)=300(ml)
Câu 2:
Ta có: \(\left\{{}\begin{matrix}n_{H_2}=\dfrac{4,48}{22,4}=0,2\left(mol\right)\\n_{HCl}=0,4\cdot2=0,8\left(mol\right)\end{matrix}\right.\)
PTHH: \(Ca+2HCl\rightarrow CaCl_2+H_2\uparrow\)
0,2____0,4_____0,2____0,2 (mol)
\(CaO+2HCl\rightarrow CaCl_2+H_2O\)
0,2____0,4______0,2____0,2 (mol)
Ta có: \(\left\{{}\begin{matrix}\%m_{Ca}=\dfrac{0,2\cdot40}{0,2\cdot40+0,2\cdot56}\cdot100\%\approx41,67\%\\\%m_{CaO}=58,33\%\\m_{CaCl_2}=\left(0,2+0,2\right)\cdot111=44,4\left(g\right)\end{matrix}\right.\)
Sửa đề: đktc → đkc
a, \(Mg+2HCl\rightarrow MgCl_2+H_2\)
\(Fe+2HCl\rightarrow FeCl_2+H_2\)
Ta có: 24nMg + 56nFe = 13,2 (1)
\(n_{H_2}=\dfrac{8,6765}{24,79}=0,35\left(mol\right)\)
Theo PT: \(n_{H_2}=n_{Mg}+n_{Fe}=0,35\left(2\right)\)
Từ (1) và (2) \(\Rightarrow\left\{{}\begin{matrix}n_{Mg}=0,2\left(mol\right)\\n_{Fe}=0,15\left(mol\right)\end{matrix}\right.\)
\(\Rightarrow\left\{{}\begin{matrix}\%m_{Mg}=\dfrac{0,2.24}{13,2}.100\%\approx36,36\%\\\%m_{Fe}\approx63,64\%\end{matrix}\right.\)
b, Theo PT: \(\left\{{}\begin{matrix}n_{MgCl_2}=n_{Mg}=0,2\left(mol\right)\\n_{FeCl_2}=n_{Fe}=0,15\left(mol\right)\end{matrix}\right.\)
⇒ m muối khan = 0,2.95 + 0,15.127 = 38,05 (g)
nH2= 0,35(mol)
a) PTHH: Mg + 2 HCl -> MgCl2 + H2
x_________2x_______x______x(mol)
PTHH: Fe + 2 HCl -> FeCl2 + H2
y________2y________y_____y(mol)
Ta có hpt: \(\left\{{}\begin{matrix}24x+56y=13,2\\x+y=0,35\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=0,2\\y=0,15\end{matrix}\right.\)
b) m=m(muối khan)= mMgCl2 + mFeCl2= 95.x+127y=95.0,2+127.0,15= 38,05(g)
a)
Gọi
\(n_{Fe} = a(mol) ; n_{Mg} = b(mol)\\ \Rightarrow 56a + 24b = 13,2(1)\)
\(Mg + 2HCl \to MgCl_2 + H_2\\ Fe + 2HCl \to FeCl_2 + H_2\)
Theo PTHH : \(n_{H_2} = a + b = 0,35(mol)\)(2)
Từ (1)(2) suy ra a = 0,15 ;b = 0,2
Vậy :
\(\%m_{Fe} = \dfrac{0,15.56}{13,2}.100\% = 63,64\%\\ \Rightarrow m_{Mg} = 100\% - 63,64\% = 36,36\%\)
b)
Ta có :\(n_{HCl} = 2n_{H_2} = 0,7(mol)\)
Bảo toàn khối lượng :
\(m_{muối} = m_{kim\ loại} + m_{HCl} - m_{H_2} = 13,2 + 0,7.36,5 - 0,35.2=38,05(gam)\)
- Thấy Cu không phản ứng với HCl .
\(\Rightarrow m_{cr}=m_{Cu}=6,4\left(g\right)\)
\(2Al+6HCl\rightarrow2AlCl_3+3H_2\)
.x.......................................1,5x.........
\(Fe+2HCl\rightarrow FeCl_2+H_2\)
.y....................................y.............
Theo bài ra ta có hệ : \(\left\{{}\begin{matrix}27x+56y+6,4=17,4\\1,5x+y=0,4\end{matrix}\right.\)
\(\Rightarrow\left\{{}\begin{matrix}x=0,2\\y=0,1\end{matrix}\right.\) ( mol )
\(\Rightarrow\left\{{}\begin{matrix}m_{Al}=5,4\\m_{Fe}=5,6\end{matrix}\right.\) ( g )
b, \(FeCl_2+2NaOH\rightarrow Fe\left(OH\right)_2+2NaCl\)
.......0,1.........0,2...............................
\(AlCl_3+3NaOH\rightarrow Al\left(OH\right)_3+3NaCl\)
...0,2.......0,6..........................
\(\Rightarrow n_{NaOH}=0,2+0,6=0,8< 1\)
=> Trong B còn có HCl dư .
\(NaOH+HCl\rightarrow NaCl+H_2O\)
...0,2..........0,2....................
=> Dư 0,2 mol HCl .
\(\Rightarrow n_{HCl}=2n_{H_2}+0,2=1\left(mol\right)\)
\(\Rightarrow m_{ddB}=17,4+250-6,4-0,8=260,2\left(g\right)\)
\(\Rightarrow\left\{{}\begin{matrix}C\%_{HCl}=\dfrac{0,2.36,5}{260,2}.100\%\approx2,8\%\\C\%_{FeCl_2}\approx4,88\%\\C\%_{AlCl_3}\approx10,26\%\end{matrix}\right.\)
Vậy ....
Câu 2.
a)
\(2Al+6HCl\rightarrow2AlCl_3+3H_2\)
\(n_{H2}=\frac{6,72}{22,4}=0,3\left(mol\right)\)
\(\rightarrow n_{Al}=0,2\left(mol\right)\)
\(m_{Al}=0,2.27=5,4\left(g\right)\)
\(m_{Cu}=11,9-5,4=6,4\left(g\right)\)
b)
\(n_{HCl}=2n_{H2}=0,6\left(mol\right)\)
\(m_{ddHCl}=\frac{0,6.36,5}{18,25\%}=120\left(g\right)\)
Câu 3:
\(n_{H2}=\frac{2,24}{22,4}=0,1\left(mol\right)\)
\(Mg+2HCl\rightarrow MgCl_2+H_2\)
0,1_____0,2____________0,1
\(MgO+2HCl\rightarrow MCl_2+H_2O\)
0,1_____0,2____________
\(m_{Mg}=0,1.24=2,4\left(g\right)\)
\(m_{MgO}=6,4-2,4=4\left(g\right)\)
\(n_{MgO}=\frac{4}{24+16}=0,1\left(mol\right)\)
\(m_{dd_{HCl}}=\frac{0,4.36,5}{20\%}=73\left(g\right)\)
\(V_{dd_{HCl}}=\frac{73}{1,2}=60,83\left(l\right)\)
Cho 78,3 gam MnO2 tác dụng vừa đủ với dd HCl 20%.
a/. Tính khối lượng dd HCl pư và thể tích khí bay ra (đkc).
b/. Khí sinh ra cho tác dụng với 250 ml dd NaOH ở t0 thường. Tính CM của dd NaOHpư và của dd muối thu được?
Bài 1
MnO2+4HCl-->MnCl2+2H2O+Cl2
a) n MNO2=78,3/87=0,9(mol)
n HCl=4n MnO2=3,6(mol)
m HCl=3,6.36,5=131,4(g)
m dd HCl=131,4.100/20=657(g)
n Cl2=n MnO2=0,9(mol)
V Cl2=0,9.22,4=20,16(l)
b) Cl2+2NaOH--->NaClO+NaCl+H2O
n NaOH=2n Cl2=1,8(mol)
CM NaOH=1,8/0,25=7,2(M)
6/ a) 2Al+6HCl--->2AlCl3+3H2
n H2=6,72/22,4=0,3(mol)
n Al=2/3n H2=0,2(mol)
m Al=0,2.27=5,4(g)
m Cu=11,8-5,4=6,4(g)
b) n HCl=2n H2=0,6( mol)
m HCl=0,6.36,5=21,9(g)
m dd HCl=21,9.100/18,25=120(g)
7/ Mg+2HCl--->MgCl2+H2
MgO+2HCl--->Mgcl2+H2O
n H2=2,24/22,4=0,1(mol)
n Mg=n H2=0,1(mol)
m Mg=0,1.24=2,4(g)
m MgO=6,4-2,4=4(g)
b) Theo pthh1
n HCl=2n H2=0,2(mol)
n MgO=4/40=0,1(mo)
Theo pthh2
n HCl=2n MgO=0,2(mol)
tổng n HCl=0,4(mol)
m HCl=0,4/36,5=14,6(g)
m dd HCl=14,6.100/20=73(g)
V HCl=72/1,2=60,833(ml)