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Gọi x,y lần lượt là số mol của Mg và Fe
\(PTHH:\\ Mg+2HCl\rightarrow MgCl_2+H_2\\ Fe+2HCl\rightarrow FeCl_2+H_2\\ n_{HCl}=0,14\cdot1=0,14\left(mol\right)\\ \Rightarrow n_{HCl}=2x+2y=0,14;m_{hh}=24x+56y=1,69\\ \Rightarrow\left\{{}\begin{matrix}x=0,0696875\\y=0,0003125\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}m_{Mg}=24x=1,6725\left(g\right)\\m_{Fe}=56y=0,0157\left(g\right)\end{matrix}\right.\)
a)\(Mg+2HCl\rightarrow MgCl_2+H_2\)
\(x\) \(2x\)
\(Fe+2HCl\rightarrow FeCl_2+H_2\)
\(y\) \(2y\)
b)\(n_{HCl}=0,14\cdot1=0,14mol\)
Ta có: \(\left\{{}\begin{matrix}24x+56y=1,69\\2x+2y=0,14\end{matrix}\right.\)\(\Rightarrow\left\{{}\begin{matrix}x=0,07\\y=3,125\cdot10^{-4}\end{matrix}\right.\)
\(m_{Mg}=0,07\cdot24=1,68\left(g\right)\)
\(m_{Fe}=3,125\cdot10^{-4}\cdot56=0,0175\left(g\right)\)
a) Fe + 2HCl --> FeCl2 + H2
b) \(n_{H_2}=\dfrac{3,36}{22,4}=0,15\left(mol\right)\)
PTHH: Fe + 2HCl --> FeCl2 + H2
_____0,15<-0,3<----0,15<---0,15
\(\left\{{}\begin{matrix}\%Fe=\dfrac{0,15.56}{12}.100\%=70\%\%\\\%Cu=100\%-70\%=30\%\end{matrix}\right.\)
c) mHCl = 0,3.36,5 = 10,95 (g)
=> \(m_{dd}=\dfrac{10,95.100}{10}=109,5\left(g\right)\)
d) mdd = 12 + 109,5 - 0,15.2 = 121,2 (g)
\(C\%\left(FeCl_2\right)=\dfrac{0,15.127}{121,2}.100\%=15,718\%\)
a) Fe + 2HCl --> FeCl2 + H2
FeO + 2HCl --> FeCl2 + H2O
b) \(n_{H_2}=\dfrac{11,2}{22,4}=0,5\left(mol\right)\)
PTHH: Fe + 2HCl --> FeCl2 + H2
______0,5<-1<------0,5<---0,5
=> mFe = 0,5.56 = 28 (g)
=> \(\left\{{}\begin{matrix}\%Fe=\dfrac{28}{100}.100\%=28\%\\\%FeO=100\%-28\%=72\%\end{matrix}\right.\)
c) \(n_{FeO}=\dfrac{72}{72}=1\left(mol\right)\)
PTHH: FeO + 2HCl --> FeCl2 + H2O
______1---->2
=> mHCl = (1+2).36,5 = 109,5 (g)
=> \(m_{ddHCl}=\dfrac{109,5.100}{30}=365\left(g\right)\)
=> \(V_{ddHCl}=\dfrac{365}{1,15}=317,39\left(ml\right)\)
\(Mg+2HCl\rightarrow MgCl_2+H_2\)
\(MgO+2HCl\rightarrow MgCl_2+H_2O\)
\(n_{Mg}=n_{H_2}=\dfrac{1.12}{22.4}=0.05\left(mol\right)\)
\(m_{Mg}=0.05\cdot24=1.2g\)
\(m_{MgO}=9.2-1.2=8\left(g\right)\)
a) Pt : \(Mg+2HCl\rightarrow MgCl_2+H_2|\)
1 2 1 1
0,3 0,6 0,3
\(MgO+2HCl\rightarrow MgCl_2+H_2O|\)
1 2 1 1
0,1 0,2
b) \(n_{Mg}=\dfrac{0,3.1}{1}=0,3\left(mol\right)\)
\(m_{Mg}=0,3.24=7,2\left(g\right)\)
\(m_{MgO}=11,2-7,2=4\left(g\right)\)
c) 0/0Mg = \(\dfrac{7,2.100}{11,2}=64,29\)0/0
0/0MgO = \(\dfrac{4.100}{11,2}=35,71\)0/0
d) Có : \(m_{MgO}=4\left(g\right)\)
\(n_{MgO}=\dfrac{4}{40}=0,1\left(mol\right)\)
\(n_{HCl\left(tổng\right)}=0,6+0,2=0,8\left(mol\right)\)
\(m_{HCl}=0,8.36,5=29,2\left(g\right)\)
\(C_{ddHCl}=\dfrac{29,2.100}{200}=14,6\)0/0
Chúc bạn học tốt
nH2=0,1(mol)
PTHH: Mg + 2 HCl -> MgCl2 + H2
0,1__________0,2___________0,1(mol)
MgO + 2 HCl -> MgCl2 + H2O
0,05____0,1___0,05(mol)
mMg=0,1. 24= 2,4(g) -> mMgO=4,4-2,4= 2(g) -> nMgO=0,05((mol)
b) %mMg= (2,4/4,4).100=54,545%
=> %mMgO=45,455%
c) nHCl=0,3(mol) -> mHCl=0,3.36,5=10,95(g)
=> mddHCl=(10,95.100)/7,3=150(g)
Thiếu đề thì phải