Hoà tan 5,6g Fe trong dd HCl vừa đủ, tính khối lượng muối thu được sau phản ứng?
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\(n_{Zn}=\dfrac{6,5}{65}=0,1(mol)\\ PTHH:Zn+H_2SO_4\to ZnSO_4+H_2\\ \Rightarrow n_{ZnSO_4}=n_{Zn}=0,1(mol)\\ \Rightarrow m_{ZnSO_4}=0,1.161=16,1(g)\)
![](https://rs.olm.vn/images/avt/0.png?1311)
\(n_{CuO}=\dfrac{32}{80}=0,4(mol)\\ CuO+2HCl\to CuCl_2+H_2\\ \Rightarrow n_{HCl}=0,8(mol);n_{CuCl_2}=n_{H_2}=0,4(mol)\\ a,m_{dd_{HCl}}=\dfrac{0,8.36,5}{20\%}=146(g)\\ b,m_{CuCl_2}=0,4.135=54(g)\\ c,C\%_{CuCl_2}=\dfrac{54}{32+146-0,4.2}.100\%=30,47\%\)
\(CuO + 2HCl \rightarrow CuCl_2 + H_2O\)
\(n_{CuO}= \dfrac{32}{80}= 0,4 mol\)
Theo PTHH:
\(n_{HCl}= 2n_{CuO}= 0,8 mol\)
\(\Rightarrow m_{HCl}= 0,8 . 36,5=29,2 g\)
\(\rightarrow m_{dd HCl}= \dfrac{29,2 . 100%}{20%}= 146 g\)
b) Muối tạo thành là CuCl2
Theo PTHH:
\(n_{CuCl_2}= n_{CuO}= 0,4 mol\)
\(\Rightarrow m_{CuCl_2}= 0,4 . 135= 54g\)
c)
\(m_{dd sau pư}= m_{CuO} + m_{dd HCl}= 32 + 146=178 g\)
C%= \(\dfrac{54}{178} . 100\)%= 30,337 %
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Câu 1:
\(PTHH_1:ACO_3+2HCl\rightarrow ACl_2+CO_2+H_2O\)
\(PTHH_2:BCO_3+2HCl\rightarrow BCl_2+CO_2+H_2O\)
\(\Rightarrow n_{HCl}=2n_{CO2}=0,4\left(mol\right)\)
Câu 2:
Ta có:
\(n_{H2}=\frac{2,24}{22,4}=0,1\left(mol\right)\)
Bảo toàn H : \(2HCl\rightarrow H_2\)
\(\Rightarrow n_{HCl}=2n_{H2}=0,2\left(mol\right)\)
BTKL: \(m_{muoi}=5,6+0,2.36,5-0,1.2=12,7\left(g\right)\)
![](https://rs.olm.vn/images/avt/0.png?1311)
Ta có: \(n_{Fe}=\dfrac{5,6}{56}=0,1\left(mol\right)\)
a, PT: \(2Fe+3Cl_2\underrightarrow{t^o}2FeCl_3\)
______0,1___0,15___0,1 (mol)
b, Có: \(m_{FeCl_3}=0,1.162,5=16,25\left(g\right)\)
c, \(C_{M_{FeCl_3}}=\dfrac{0,1}{0,1}=1M\)
Bạn tham khảo nhé!
![](https://rs.olm.vn/images/avt/0.png?1311)
\(a)2Al+6HCl\rightarrow2AlCl_3+3H_2\\ Fe+2HCl\rightarrow FeCl_2+H_2\\ b)n_{H_2}=\dfrac{5,6}{22,4}=0,25mol\\ n_{Al}=a;n_{Fe}=b\\ \left\{{}\begin{matrix}3a+b=0,25\\27a+56b=8,3\end{matrix}\right.\\ a=\dfrac{19}{470};b=\dfrac{121}{940}\\ \%m_{Al}=\dfrac{\dfrac{19}{470}\cdot27}{8,3}\cdot100=13,15\%\\ \%m_{Fe}=100-13,15=86,85\%\\ c)n_{HCl}=3\cdot\dfrac{19}{470}+2\cdot\dfrac{121}{940}=\dfrac{89}{235}mol\\ m_{ddHCl=}=\dfrac{\dfrac{89}{235}\cdot36,5}{7,3}\cdot100=189g\\ d)n_{AlCl_3}=n_{Al}=\dfrac{19}{470}mol\\ n_{Fe}=n_{FeCl_2}=\dfrac{121}{940}mol\)
\(m_{dd}=8,3+189-0,25.2=196,8g\\ C_{\%AlCl_3}=\dfrac{\dfrac{19}{470}\cdot133,8}{196,8}\cdot100=2,8\%\\ C_{\%FeCl_2}=\dfrac{\dfrac{121}{940}127}{196,8}\cdot100=8,3\%\)
Bảo toàn nguyên tố Fe:
\(n_{FeCl_2}=n_{Fe}=0,1\left(mol\right)\)
\(\Rightarrow m_{FeCl_2}=12,7\left(g\right)\)