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\(\text{Ta có PTHH}\\2Al+6HCl \rightarrow 2AlCl_3+3H_2 \uparrow\\n_{Al}=\dfrac{5,4}{27}=0,2(mol)\\\Rightarrow n_{HCl}=3n_{Al}=0,6(mol)\\\Rightarrow C_{M_{HCl}}=n/V=\dfrac{0,6}{0,15}=4(M)\\\text{ Câu hỏi 1 : B}\\\Rightarrow n_{AlCl_3}=n_{Al}=0,2(mol)\\\Rightarrow m_{AlCl_3} = 0,2.133,5=26,7(gam)\\\text{ Câu hỏi 2 : A}\\\Rightarrow n_{H_2}=3/2n_{Al}=0,3(mol)\Rightarrow V_{H_2}(đktc)=0,3.22,4=6,72(lít)\\\text{ Câu hỏi 3 : C} \)
a. PTHH: \(2Al+6HCl\rightarrow2AlCl_3+3H_2\uparrow\)
\(n_{H_2}=\frac{6,72}{22,4}=0,3mol\)
Theo phương trình \(n_{Al}=n_{AlCl_3}=\frac{2}{3}n_{H_2}=0,2mol\)
\(\rightarrow m_{Al}=0,2.27=5,4g\)
\(\rightarrow m=5,4\)
b. \(m_{\text{muối}}=m_{AlCl_3}=0,2.133,5=26,7g\)
a)PTHH\(2AL+6HCL\rightarrow2ALCL_3+3H_2\uparrow\)
\(n_{H_2}=\frac{6,72}{22,4}=0,3mol\)
Theo phương trình:\(n_{AL}=n_{alcl_3}=\frac{2}{3}n_{H_2}=0,2mol\)
\(\rightarrow m_{AL}=0,2\cdot27=5,4g\)
\(\rightarrow m=5,4\)
b)\(m_{muối}=m_{alcl_3}=0,2\cdot133,5=26,7g\)
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) $n_{Fe} = \dfrac{11,2}{56} = 0,2(mol)$
$Fe^0 \to Fe^{+3} + 3e$
$N^{+5} + 3e \to N^{+2}$
Bảo toàn electron : $3n_{Fe} = 3n_{NO}$
$\Rightarrow n_{NO} = 0,2(mol)$
$\Rightarrow V = 0,2.22,4 = 4,48(lít)$
b) $n_{Fe(NO_3)_3} = n_{Fe} = 0,2(mol)$
$\Rightarrow m = 0,2.242 = 48,4(gam)$
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)
$n_{MgCl_2} = n_{Mg} = 0,2(mol) \Rightarrow m_{MgCl_2} = 0,2.95 = 19(gam)$
\(n_{Mg}=\dfrac{4,8}{24}=0,2\left(mol\right)\\ Mg+2HCl\rightarrow MgCl_2+H_2\\ 0,2.......0,4........0,2.........0,2\left(mol\right)\\ a.m_{HCl}=0,4.36,5=14,6\left(g\right)\\ b.m_{MgCl_2}=0,2.95=19\left(g\right)\)
\(2Al+6HCl->2AlCl_3+3H_2\\ Fe+2HCl->FeCl_2+H_2\\ n_{Al}=a;n_{Fe}=b\\ 27a+56b=8,3\\ 1,5a+b=\dfrac{5,6}{22,4}=0,25\\ a=b=0,1\\ m_{Al}=27\cdot0,1=2,7g\\ m_{Fe}=8,3-2,7=5,6g\\ a=\dfrac{3a+2b}{500}\cdot36,5=3,65\%\\ m_{ddsau}=508,3-0,25\cdot2=507,8g\\ C\%_{AlCl_3}=\dfrac{133,5a}{507,8}=2,63\%\\ C\%_{FeCl_2}=\dfrac{127b}{507,8}=2,50\%\)
2al+6hcl-> 2alcl3+ 3h2
fe+2hcl-> fecl2+h2
nh2=13,44/22,4=0,6 mol
27a+56b=16,5
1,5a+ b=0,6
a=0,3, b=0,15
%mal=0,3*27/16,5*100=49,09%
%mfe=50,9%
nhcl=3a+2b=1,2
Vdd hcl=1,2/2=0,6l
\(a,Fe+2HCl\rightarrow FeCl_2+H_2\\ CuO+2HCl\rightarrow CuCl_2+H_2O\\ n_{H_2}=\dfrac{2,24}{22,4}=0,1\left(mol\right)\Rightarrow n_{Fe}=n_{H_2}=0,1\left(mol\right)\\ \Rightarrow\%m_{Fe}=\dfrac{0,1.56}{13,6}.100\%\approx41,176\%\\ \Rightarrow\%m_{CuO}\approx58,824\%\\ b,n_{CuO}=\dfrac{13,6-0,1.56}{80}=0,1\left(mol\right)\\ n_{HCl\left(p.ứ\right)}=2.\left(n_{Fe}+n_{CuO}\right)=2.\left(0,1+0,1\right)=0,4\left(mol\right)\\ \Rightarrow V_{ddHCl}=\dfrac{0,4}{2}=0,2\left(l\right)\)
Zn + 2HCl => ZnCl2 + H2
nZn = m/M = 6.5/65 = 0.1 (mol)
==> nHCl = 0.2 (mol), nZnCl2 = 0.1 (mol) = nH2
mZnCl2 = n.M = 135x0.1 = 13.6 (g)
VH2 = 22.4xn = 22.4 x 0.1 = 2.24 (l)
CM dd HCl = n/V = 0.2/0.5 = 0.4 M
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!