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![](https://rs.olm.vn/images/avt/0.png?1311)
\(a) m_{Cu} = 9,6(gam)\\ n_{Al} = a(mol) ; n_{Fe} = b(mol)\\ \Rightarrow 27a + 56b = 16,55 -9,6 =6,95(1)\\ 2Al + 6HCl \to 2AlCl_3 + 3H_2\\ Fe + 2HCl \to FeCl_2 + H_2\\ n_{H_2} = 1,5a + b = \dfrac{3,92}{22,4} = 0,175(2)\\ (1)(2) \Rightarrow a = 0,05 ; b = 0,1\\ m_{Al} = 0,05.27 = 1,35(gam); n_{Fe} = 0,1.56 = 5,6(gam)\)
\(b) n_{HCl} = 2n_{H_2} = 0,175.2 = 0,35(mol) \Rightarrow m_{HCl} = 0,35.36,5 = 12,775(gam)\)
![](https://rs.olm.vn/images/avt/0.png?1311)
\(n_{H_2\left(dktc\right)}=\dfrac{V}{22,4}=\dfrac{2,464}{22,4}=0,11\left(mol\right)\)
Đặt \(\left\{{}\begin{matrix}n_{Ba}=a\left(mol\right)\\n_K=b\left(mol\right)\end{matrix}\right.\left(a,b>0\right)\)
\(PTHH:Ba+2H_2O->Ba\left(OH\right)_2+H_2\left(1\right)\)
tỉ lệ 1 ; 2 : 1 : 1
n(mol) a--------->2a----------->a---------->a
\(PTHH:2K+2H_2O->2KOH+H_2\left(2\right)\)
tỉ lệ 2 : 2 : 2 ; 1
n(mol) b---------->b---------->b------------>1/2b
Ta có Hệ phương trình sau
\(\left\{{}\begin{matrix}137a+39b=11,53\\a+\dfrac{1}{2}b=0,11\end{matrix}\right.\\ < =>\left\{{}\begin{matrix}a=0,05\\b=0,12\end{matrix}\right.\\ < =>\left\{{}\begin{matrix}n_{Ba}=0,05\left(mol\right)\\n_K=0,12\left(mol\right)\end{matrix}\right.\)
Theo Phương trình (1) ta có: \(n_{Ba\left(OH\right)_2}=a=0,05\left(mol\right)\\ =>m_{Ba\left(OH\right)_2}=n\cdot M=0,05\cdot171=8,55\left(g\right)\)
Theo phương trình (2) ta có
\(n_{KOH}=b=0,12\left(mol\right)\\ m_{KOH}=n\cdot M=0,12\cdot56=6,72\left(g\right)\\ =>m_{ct}=8,55+6,72=15,27\left(g\right)\)
![](https://rs.olm.vn/images/avt/0.png?1311)
2Al + 3H2SO4 → Al2(SO4)3 + 3H2
\(n_{H_2}=\frac{10,08}{22,4}=0,45\left(mol\right)\)
a) Theo pt: \(n_{Al}=\frac{2}{3}n_{H_2}=\frac{2}{3}\times0,45=0,3\left(mol\right)\)
\(\Rightarrow m_{Al}=0,3\times27=8,1\left(g\right)\)
\(\Rightarrow m_{Cu}=14,5-8,1=6,4\left(g\right)\)
\(\%m_{Al}=\frac{8,1}{14,5}\times100\%=55,86\%\)
\(\%m_{Cu}=\frac{6,4}{14,5}\times100\%=44,14\%\)
b) Theo pT: \(n_{H_2SO_4}=n_{H_2}=0,45\left(mol\right)\)
\(\Rightarrow m_{H_2SO_4}=0,45\times98=44,1\left(g\right)\)
\(\Rightarrow m_{ddH_2SO_4}=\frac{44,1}{20\%}=220,5\left(g\right)\)
c) \(m_{H_2}=0,45\times2=0,9\left(g\right)\)
Ta có: \(m_{dd}saupứ=8,1+220,5-0,9=227,7\left(g\right)\)
Theo pT: \(n_{Al_2\left(SO_4\right)_3}=\frac{1}{3}n_{H_2}=\frac{1}{3}\times0,45=0,15\left(mol\right)\)
\(\Rightarrow m_{Al_2\left(SO_4\right)_3}=0,15\times342=51,3\left(g\right)\)
\(\Rightarrow C\%_{Al_2\left(SO_4\right)_3}=\frac{51,3}{227,7}\times100\%=22,53\%\)
Gọi số mol của Al và Cu lần lượt là x và y
Vì Cu k phản ứng với dd H2SO4 20% ( đã bị pha loãng) nên ta có
\(PTHH:2Al+3H_2SO_4\rightarrow Al_2\left(SO_4\right)_3+3H_2\)
(mol) 2 3 1 3
(mol) x 3x/2 x/2 3x/2
Theo đề bài ta có:
\(hpt:\left\{{}\begin{matrix}22,4\times\frac{3x}{2}=10,08\\27x+64y=14,5\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=0,3\\y=0,1\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}n_{Al}=0,3\left(mol\right)\rightarrow\%m_{Al}=\frac{0,3.27}{14,5}.100\%=55,9\left(\%\right)\\n_{Cu}=0,1\left(mol\right)\rightarrow\%m_{Cu}=100-55,9=44,1\left(\%\right)\end{matrix}\right.\)
\(m_{H_2SO_4}=n.M=\frac{98.3x}{2}=98.\frac{3.0,3}{2}=44,1\left(g\right)\)
\(m_{ddH_2SO_4}=\frac{44,1.100\%}{20\%}=220,5\left(g\right)\)
\(C\%_{ddAl_2\left(SO_4\right)_3}=\frac{m_{ct}}{m_{dd}}.100\%=\frac{342.\frac{0,3}{2}}{220,5+14,5-2.\frac{3.0,3}{2}}.100\%=21,91\left(\%\right)\)
![](https://rs.olm.vn/images/avt/0.png?1311)
`a)Fe + 2HCl -> FeCl_2 + H_2 \uparrow`
`0,1` `0,1` `(mol)`
`Cu + HCl -xx->`
`b)n_[H_2]=[2,479]/[22,4]=0,1 (mol)`
`m_[Fe]=0,1.56=5,6(g)`
`=>m_[Cu]=10-5,6=4,4(g)`
`c)%m_[Fe]=[5,6]/10 .100=56%`
`%m_[Cu]=100-56=44%`
`d)` Dung dịch sau phản ứng có làm đổi màu quỳ tím. Vì: `HCl` dư nên sau phản ứng quỳ tím đổi màu đỏ.
![](https://rs.olm.vn/images/avt/0.png?1311)
Chất rắn D là Cu, chất rắn E là CuO
\(m_{tăng}=m_{O_2}=0,16\left(g\right)\)
=> \(n_{O_2}=\dfrac{0,16}{32}=0,005\left(mol\right)\)
PTHH: 2Cu + O2 --to--> 2CuO
0,01<-0,005
=> mCu = 0,01.64 = 0,64 (g)
Gọi số mol K, Ba là a, b (mol)
=> 39a + 137b = 3,18 - 0,64 = 2,54 (1)
PTHH: 2K + 2H2O --> 2KOH + H2
a--------------->a
Ba + 2H2O --> Ba(OH)2 + H2
b--------------->b
=> 56a + 171b = 3,39 (2)
(1)(2) => a = 0,03 (mol); b = 0,01 (mol)
=> \(\left\{{}\begin{matrix}\%m_{Cu}=\dfrac{0,64}{3,18}.100\%=20,126\%\\\%m_K=\dfrac{0,03.,39}{3,18}.100\%=36,792\%\\\%m_{Ba}=\dfrac{0,01.137}{3,18}.100\%=43,082\%\end{matrix}\right.\)
\(m_{O_2}=m+0,16-m=0,16\left(g\right)\\ \rightarrow n_{O_2}=\dfrac{0,16}{32}=0,005\left(mol\right)\)
PTHH: 2Cu + O2 --to--> 2CuO
0,01 0,005
Gọi \(\left\{{}\begin{matrix}n_K=a\left(mol\right)\\n_{Ba}=b\left(mol\right)\end{matrix}\right.\)
PTHH:
2K + 2H2O ---> 2KOH + H2
a a
Ba + 2H2O ---> Ba(OH)2 + H2
b b
Hệ pt \(\left\{{}\begin{matrix}39a+137b=3,18-0,01.64=2,54\\56a+171b=3,39\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}a=0,03\left(mol\right)\\b=0,01\left(mol\right)\end{matrix}\right.\)
\(\rightarrow\left\{{}\begin{matrix}\%m_{Cu}=\dfrac{0,01.64}{3,18}=20,13\%\\\%m_K=\dfrac{0,03.39}{3,18}=36,79\%\\\%m_{Ba}=100\%-20,13\%-36,79\%=43,08\%\end{matrix}\right.\)
![](https://rs.olm.vn/images/avt/0.png?1311)
1.Gọi công thức tổng quát của 2 muối cacbonat đó là: MCO3, N2(CO3)3
MCO3(x)+2HCl(2x)→MCl2(x)+H2O+CO2(x)MCO3(x)+2HCl(2x)→MCl2(x)+H2O+CO2(x)
N2(CO3)3(y)+6HCl(6y)→2NCl3(2y)+3H2O+3CO2(3y)N2(CO3)3(y)+6HCl(6y)→2NCl3(2y)+3H2O+3CO2(3y)
Gọi số mol MCO3 và N2(CO3)3 lần lược là x, y ta có
(M+60)x+(2N+180)y=3,34(M+60)x+(2N+180)y=3,34
⇔Mx+2Ny+60(x+3y)=3,34(1)⇔Mx+2Ny+60(x+3y)=3,34(1)
Ta lại có: nCO2=0,89622,4=0,04nCO2=0,89622,4=0,04
⇒x+3y=0,04(2)⇒x+3y=0,04(2)
Thế (2) vào (1) ta được: Mx+2Ny+60.0,04=3,34Mx+2Ny+60.0,04=3,34
⇔Mx+2Ny=0,94(3)⇔Mx+2Ny=0,94(3)
Ta cần tính: mhhm=(M+71)x+(N+106,5).2ymhhm=(M+71)x+(N+106,5).2y
=Mx+2Ny+71(x+3y)=0,94+71.0,04=3,78
![](https://rs.olm.vn/images/avt/0.png?1311)
a) Do dd sau pư có 3 chát tan với nồng độ % bằng nhau
=> \(m_{Al_2\left(SO_4\right)_3}=m_{ZnSO_4}=m_{H_2SO_4\left(dư\right)}\)
Gọi số mol Al, Zn là a, b (mol)
PTHH: 2Al + 3H2SO4 --> Al2(SO4)3 + 3H2
a----->1,5a------->0,5a----->1,5a
Zn + H2SO4 --> ZnSO4 + H2
b----->b--------->b----->b
=> \(\left\{{}\begin{matrix}m_{Al_2\left(SO_4\right)_3}=342.0,5a=171a\left(g\right)\\m_{ZnSO_4}=161b\left(g\right)\end{matrix}\right.\)
=> 171a = 161b
=> \(\dfrac{a}{b}=\dfrac{161}{171}\) (1)
Có: \(\dfrac{m_{Al}}{m_{Zn}}=\dfrac{27.n_{Al}}{65.n_{Zn}}=\dfrac{27}{65}.\dfrac{161}{171}=\dfrac{483}{1235}\)
b) \(n_{H_2}=1,5a+b=\dfrac{11,2}{22,4}=0,5\left(mol\right)\) (2)
(1)(2) => \(\left\{{}\begin{matrix}a=\dfrac{161}{825}\left(mol\right)\\b=\dfrac{57}{275}\left(mol\right)\end{matrix}\right.\)
=> \(x=\dfrac{161}{825}.27+\dfrac{57}{275}.65=\dfrac{5154}{275}\left(g\right)\)
\(m_{H_2SO_4\left(dư\right)}=m_{Al_2\left(SO_4\right)_3}=342.0,5\dfrac{161}{825}=\dfrac{9177}{275}\left(g\right)\)
=> \(m_{H_2SO_4\left(bđ\right)}=98\left(1,5a+b\right)+\dfrac{9177}{275}=\dfrac{22652}{275}\left(g\right)\)
=> \(y=\dfrac{\dfrac{22652}{275}.100}{10}=\dfrac{45304}{55}\left(g\right)\)
a/ Chỉ có Al tan , Ag và Cu không t/d
PTHH :2Al + 3H2SO4 \(\rightarrow\) Al2(SO4)3 + 3H2
mol :0,24....0,36...........0,12.............0,36
b/ Có : nH2 = 8,064/22,4 = 0,36(mol)
mAl = 0,24 . 27 =6,48(g)
=> m(Cu + Ag) = 14 - 6,48 = 7,52(g)
Đặt mAg = a (g) => mCu = 3a(g)
=> a + 3a = 7,52
=> a = 1,88(g)
Do đó : \(\left\{{}\begin{matrix}m_{Al}=6,48\left(g\right)\\m_{Ag}=a=1,88\left(g\right)\\m_{Cu}=3a=3.1,88=5,64\left(g\right)\end{matrix}\right.\)
c/ mH2SO4 (Pứ) = 0,36 . 98 =35,28(g)
=> mdd H2SO4 (Pứ)= \(\dfrac{35,28.100\%}{24,5\%}=144\left(g\right)\)
mà cần dùng dư 15 % so với lý thuyết
=> mdd H2SO4 (cần dùng) = 144 + 15%. 144 =165,6(g)
d/ Vì t/d với dd H2SO4 loãng dư
=> dd D gồm \(\left\{{}\begin{matrix}H_2SO_4dư\\Al_2\left(SO_4\right)_3\end{matrix}\right.\)
Theo ĐLBTKL :
mdd sau pứ = mAl + mdd H2SO4 - mH2 = 6,48 + 165,6 - 0,36.2 =171,36(g)
=> C%Al2(SO4)3 / dd D = \(\dfrac{0,12.342}{171,36}.100\%=23,95\%\)
Do dùng dư 15% so với lý thyết
=> mH2SO4 (dư ) = 15% . 35,28 =5,292(g)
=> C%H2SO4 / dd D = \(\dfrac{5,292}{171,36}.100\%=3,1\%\)
\(n_{H2} = \dfrac{8,064}{22,4} = 0,36 (mol)\)
( \(H_{2}SO_{4}\) loãng nên Ag, Cu không tác dụng được )
Pt : \(2Al + 3H_{2}SO_{4} \rightarrow Al_{2}(SO_{4}) _{3} + 3H_{2}\)
2 mol 3 mol 3 mol
0,24 mol<--- 0,36 mol <--- 0,36 mol
\(\Rightarrow m_{Al} = 0,24 .27 =6,48 (g)\)
Theo gt: \(m_{Cu} = 3m_{Ag}\)
\(\Rightarrow m_{Ag} + m_{Cu} + m_{Al} = 14\)
\(\Leftrightarrow m_{Ag} + 3m_{Ag} + 6,48 = 14\)
\(\Leftrightarrow 4m_{Ag} = 7,52 \)
\(\Leftrightarrow m_{Ag} = 1,88 (g)\)
\(\Rightarrow m_{Cu} = 3m_{Ag} = 3.1,88 =5,64 (g)\)
\(m_{H_{2}SO_{4}} = \dfrac{0,36.98.100}{24,5} = 144(g)\) (1 )
Vì dùng dư 15% => 100 + 15 = 1,15 (2)
(1)(2) => \(m_{{dd}_{H_{2}SO_{4}}} = 144.1,15 = 165,6 (g)\)
\(m_{dd\left(spu\right)}=m_{Al}+m_{H_2SO_4}-m_{H_2}\)
= 6,48 + 165,6 - 0,36.2
= 171,36 (g)
\(C\%_{Al_2\left(SO_4\right)_3}=\dfrac{0,36.342.100}{171,36}=20,58\%\)