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![](https://rs.olm.vn/images/avt/0.png?1311)
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: \(\left\{{}\begin{matrix}\%m_{Ca}=\dfrac{0,2\cdot40}{0,2\cdot40+0,2\cdot56}\cdot100\%\approx41,67\%\\\%m_{CaO}=58,33\%\\m_{CaCl_2}=\left(0,2+0,2\right)\cdot111=44,4\left(g\right)\end{matrix}\right.\)
![](https://rs.olm.vn/images/avt/0.png?1311)
Ta có: nMgCl2=\(\dfrac{38}{95}\)=0,4(mol);
nCO2=\(\dfrac{6,72}{22,4}\)=0,3(mol)
MgCO3 + 2HCl → MgCl2 + CO2 + H2O
(mol) 0,3 ← 0,3 ← 0,3
MgO + 2HCl → MgCl2 + H2O
(mol) 0,1 ← 0,1
\(\left\{{}\begin{matrix}mMgO=0,1.40=4g\\mMgCO3=0,3.84=25,2\end{matrix}\right.\)
=>%mMgO=\(\dfrac{4}{29,2}\).100=13,7%
=>%m MgCO4=86,3%
![](https://rs.olm.vn/images/avt/0.png?1311)
a)Gọi x,y lần lượt là số mol của Al, Fe trong hỗn hợp ban đầu (x,y>0)
Sau phản ứng hỗn hợp muối khan gồm: \(\left\{{}\begin{matrix}AlCl_3:x\left(mol\right)\\FeCl_2:y\left(mol\right)\end{matrix}\right.\)
Ta có hệ phương trình: \(\left\{{}\begin{matrix}27x+56y=13,9\\133,5x+127y=38\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x\approx0,0896\\y\approx0,205\end{matrix}\right.\)
\(\Rightarrow\left\{{}\begin{matrix}\%m_{Al}=\dfrac{0,0896\cdot27\cdot100\%}{13,9}\approx17,4\%\\\%m_{Fe}=\dfrac{0,205\cdot56\cdot100\%}{13,9}\approx82,6\%\end{matrix}\right.\)
Theo Bảo toàn nguyên tố Cl, H ta có:\(n_{H_2}=\dfrac{n_{HCl}}{2}=\dfrac{3n_{AlCl_3}+2n_{FeCl_2}}{2}\\ =\dfrac{3\cdot0,0896+2\cdot0,205}{2}=0,3394mol\\ \Rightarrow V_{H_2}=0,3394\cdot22,4\approx7,6l\)
![](https://rs.olm.vn/images/avt/0.png?1311)
\(a)\\ Zn + 2HCl \to ZnCl_2 + H_2\\ CuO + 2HCl \to CuCl_2 + H_2O\\ n_{Zn} = n_{H_2} = \dfrac{2,24}{22,4} = 0,1(mol)\\ \Rightarrow \%m_{Zn} = \dfrac{0,1.65}{22,5}.100\% = 28,89\%\\ \%m_{CuO} = 100\% - 28,89\% = 71,11\%\\ b)\ n_{CuO} = 0,2\ mol\\ n_{HCl\ pư} = 2n_{Zn} + 2n_{CuO} = 0,6(mol)\\ n_{ZnCl_2} = n_{Zn} = 0,1(mol);n_{CuCl_2} = n_{CuO} = 0,2(mol)\\ \Rightarrow C_{M_{HCl\ dư}} = \dfrac{0,5.2-0,6}{0,5}=0,8M\\ \)
\(C_{M_{ZnCl_2}} = \dfrac{0,1}{0,5} = 0,2M\\ C_{M_{CuCl_2}} = \dfrac{0,2}{0,5} = 0,4M\)
![](https://rs.olm.vn/images/avt/0.png?1311)
Gọi công thức chung của 2 kim loại là R
\(HCIII\text{R + 2HCl -> RCl2 + H2}I->RCI2+H2\)
Ta có : \(nH2=0,3mol->M\text{ R}=8,8\)/\(0,3=29,3\)
Ta có : \(\text{24 < 29,3 < 40 nên 2 kim loại là Mg và Ca}\)
Gọi số mol Mg và Ca lần lượt là x , y
\(\text{-> x+y=0,3; 24x+40y=8,8}\)
Giải được \(\text{x = 0,2 ; y = 0,1 }\)
\(\text{-> mMg=24.0,2=4,8 gam -> %Mg=54,5% -> %Ca=45,5%}\)
![](https://rs.olm.vn/images/avt/0.png?1311)
nHCl=0,7.1=0,7(mol)
nCO2=4,48/22,4=0,2(mol)
PTHH: MgCO3 +2 HCl -> MgCl2 + CO2 + H2O
0,2_________0,4_____0,2______0,2(mol)
nMgCO3=nCO2=0,2(mol) => mMgCO3=0,2. 84= 16,8(g)
=> mFeO= mX - mMgCO3= 24 - 16,8= 7,2(g)
=> %mMgCO3= (16,8/24).100=70%
=>%mFeO=100% - 70%= 30%
b) nFeO= 7,2/72=0,1(mol)
FeO +2 HCl -> FeCl2 + H2O
0,1___0,2______0,1(mol)
Vì 0,4+0,2=0,6 => HCl có dư => nHCl(dư)= 0,7 - 0,6=0,1(mol)
Vddsau= VddHCl=0,7(l)
CMddHCl(dư)= 0,1/0,7= 1/7 (M)
CMddFeCl2= 0,1/0,7=1/7(M)
CMddMgCl2= 0,2/0,7=2/7(M)
a. \(n_{CO_2}=0,2\left(mol\right)\)
\(MgCO_3+2HCl\rightarrow MgCl_2+H_2O+CO_2\)
\(\Rightarrow n_{MgCl_2}=n_{MgCO_3}=n_{CO_2}=0.2\left(mol\right)\)
\(\Rightarrow m_{MgCO_3}=0,2.84=16,8\left(g\right)\)
\(\%m_{MgCO_3}=\dfrac{16,8.100\%}{24}=70\%\\ \%m_{FeO}=100\%-70\%=30\%\)
b. \(FeO+2HCl\rightarrow FeCl_2+H_2O\)
\(n_{FeO}=\dfrac{24-16,8}{72}=0,1\left(mol\right)=n_{FeCl_2}\)
\(C_{M_{FeCl_2}}=\dfrac{0,1}{0,7}=\dfrac{1}{7}\left(M\right)\)
\(C_{M_{MgCO_3}}=\dfrac{0,2}{0,7}=\dfrac{2}{7}\left(M\right)\)
![](https://rs.olm.vn/images/avt/0.png?1311)
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 :
\(m_{muối} = m_{kim\ loại} + m_{HCl} - m_{H_2} = 13,2 + 0,7.36,5 - 0,35.2=38,05(gam)\)
![](https://rs.olm.vn/images/avt/0.png?1311)
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)
a)
\(n_{MgCl_2}=\dfrac{38}{95}=0,4\left(mol\right)\)
\(n_{CO_2}=\dfrac{6,72}{22,4}=0,3\left(mol\right)\)
PTHH: MgCO3 + 2HCl --> MgCl2 + CO2 + H2O
0,3<------0,6<------0,3<----0,3
MgO + 2HCl --> MgCl2 + H2O
0,1<---0,2<------0,1
=> \(\left\{{}\begin{matrix}m_{MgO}=0,1.40=4\left(g\right)\\m_{MgCO_3}=0,3.84=25,2\left(g\right)\end{matrix}\right.\)
b) \(m_{HCl}=\left(0,6+0,2\right).36,5=29,2\left(g\right)\)
=> \(m_{dd.HCl}=\dfrac{29,2.100}{20}=146\left(g\right)\)