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Fe+2HCl->FeCl2+H2
x---2x-----------x
Mg+2HCl->MgCl2+H2
y------2y-----------y
Ta có :
\(\left\{{}\begin{matrix}56x+24y=24\\x+y=\dfrac{13,44}{22,4}\end{matrix}\right.\)
=>x=0,3 mol, y=0,3 mol
=>%m Fe=\(\dfrac{0,3.56}{24}.100\)=70%
=>%m Mg=100-70=30%
=>VHCl=\(\dfrac{0,3.2+0,3.2}{2}\)=0,6l=600ml
b)
XCl2+2AgNO3->2AgCl+X(NO3)2
0,6--------------------1,2mol
=>m AgCl=1,2.143,5=172,2g
a) $n_{Mg} = \dfrac{4,8}{24} = 0,2(mol)$
$Mg + 2HCl \to MgCl_2 + H_2$
Theo PTHH :
$n_{HCl} = 2n_{Mg} = 0,4(mol) \Rightarrow m_{HCl} = 0,4.36,5 = 14,6(gam)$
b)
$n_{MgCl_2} = n_{Mg} = 0,2(mol) \Rightarrow m_{MgCl_2} = 0,2.95 = 19(gam)$
\(n_{Mg}=\dfrac{4,8}{24}=0,2\left(mol\right)\\ Mg+2HCl\rightarrow MgCl_2+H_2\\ 0,2.......0,4........0,2.........0,2\left(mol\right)\\ a.m_{HCl}=0,4.36,5=14,6\left(g\right)\\ b.m_{MgCl_2}=0,2.95=19\left(g\right)\)
\(\text{Ta có PTHH}\\2Al+6HCl \rightarrow 2AlCl_3+3H_2 \uparrow\\n_{Al}=\dfrac{5,4}{27}=0,2(mol)\\\Rightarrow n_{HCl}=3n_{Al}=0,6(mol)\\\Rightarrow C_{M_{HCl}}=n/V=\dfrac{0,6}{0,15}=4(M)\\\text{ Câu hỏi 1 : B}\\\Rightarrow n_{AlCl_3}=n_{Al}=0,2(mol)\\\Rightarrow m_{AlCl_3} = 0,2.133,5=26,7(gam)\\\text{ Câu hỏi 2 : A}\\\Rightarrow n_{H_2}=3/2n_{Al}=0,3(mol)\Rightarrow V_{H_2}(đktc)=0,3.22,4=6,72(lít)\\\text{ Câu hỏi 3 : C} \)
Bài 1:
\(n_{HCl}=2.0,16=0,32\left(mol\right);n_{H_2}=\dfrac{3,584}{22,4}=0,16\left(mol\right)\)
PTHH: Mg + 2HCl → MgCl2 + H2
PTHH: Fe + 2HCl → FeCl2 + H2
\(m_{H_2}=0,16.2=0,32\left(g\right)\)
\(m_{HCl}=0,32.36,5=11,68\left(g\right)\)
Theo ĐLBTKL ta có: \(m_{MgCl_2+FeCl_2}=1,4+11,68-0,32=12,76\left(g\right)\)
Bài 12:
Theo ĐLBTKL, ta có:
\(m_{hhkl}+m_{O_2}=m_{hh.oxit}\\ \Leftrightarrow11,9+m_{O_2}=18,3\\ \Leftrightarrow m_{O_2}=18,3-11,9=6,4\left(g\right)\\ n_{O_2}=\dfrac{6,4}{32}=0,2\left(mol\right)\\ V_{O_2\left(đktc\right)}=0,2.22,4=4,48\left(l\right)\)
a.\(n_{H_2}=\dfrac{V_{H_2}}{22,4}=\dfrac{8,96}{22,4}=0,4mol\)
Gọi \(\left\{{}\begin{matrix}n_{Fe}=x\\n_{Zn}=y\end{matrix}\right.\)
\(Fe+2HCl\rightarrow FeCl_2+H_2\)
x x ( mol )
\(Zn+2HCl\rightarrow ZnCl_2+H_2\)
y y ( mol )
Ta có:
\(\left\{{}\begin{matrix}56x+65y=25,55\\x+y=0,4\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}x=0,05\\y=0,35\end{matrix}\right.\)
\(\Rightarrow m_{Fe}=0,05.56=2,8g\)
\(\Rightarrow m_{Zn}=0,35.65=22,75g\)
\(\%m_{Fe}=\dfrac{2,8}{25,55}.100=10,95\%\)
\(\%m_{Zn}=100\%-10,95\%=89,05\%\)
b.\(n_{HCl}=2.0,05+2.0,35=0,8mol\)
\(C_M=\dfrac{n}{V}\Rightarrow V=\dfrac{n}{C_M}=\dfrac{0,8}{2}=0,4l\)
a) $n_{Fe} = \dfrac{11,2}{56} = 0,2(mol)$
$Fe^0 \to Fe^{+3} + 3e$
$N^{+5} + 3e \to N^{+2}$
Bảo toàn electron : $3n_{Fe} = 3n_{NO}$
$\Rightarrow n_{NO} = 0,2(mol)$
$\Rightarrow V = 0,2.22,4 = 4,48(lít)$
b) $n_{Fe(NO_3)_3} = n_{Fe} = 0,2(mol)$
$\Rightarrow m = 0,2.242 = 48,4(gam)$
\(2Al+6HCl->2AlCl_3+3H_2\\ Fe+2HCl->FeCl_2+H_2\\ n_{Al}=a;n_{Fe}=b\\ 27a+56b=8,3\\ 1,5a+b=\dfrac{5,6}{22,4}=0,25\\ a=b=0,1\\ m_{Al}=27\cdot0,1=2,7g\\ m_{Fe}=8,3-2,7=5,6g\\ a=\dfrac{3a+2b}{500}\cdot36,5=3,65\%\\ m_{ddsau}=508,3-0,25\cdot2=507,8g\\ C\%_{AlCl_3}=\dfrac{133,5a}{507,8}=2,63\%\\ C\%_{FeCl_2}=\dfrac{127b}{507,8}=2,50\%\)
PT: \(Fe+2HCl\rightarrow FeCl_2+H_2\)
Ta có: \(n_{Fe}=\dfrac{11,2}{56}=0,2\left(mol\right)\)
Theo PT: \(\left\{{}\begin{matrix}n_{H_2}=n_{FeCl_2}=n_{Fe}=0,2\left(mol\right)\\n_{HCl}=2n_{Fe}=0,4\left(mol\right)\end{matrix}\right.\)
a, Ta có: \(V_{H_2}=0,2.22,4=4,48\left(l\right)\)
b, \(V_{ddHCl}=\dfrac{0,4}{1,5}\approx0,267\left(l\right)\)
c, \(m_{FeCl_2}=0,2.127=25,4\left(g\right)\)
Bạn tham khảo nhé!