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a) CaCO3 + 2HCl --> CaCl2 + CO2 + H2O
b) \(n_{CaCO_3}=\dfrac{10}{100}=0,1\left(mol\right)\)
CaCO3 + 2HCl --> CaCl2 + CO2 + H2O
_0,1---->0,2------->0,1----->0,1
=> mCaCl2 = 0,1.111 = 11,1 (g)
=> VCO2 = 0,1.22,4 = 2,24 (l)
c) \(a=C_{M\left(HCl\right)}=\dfrac{0,2}{0,4}=0,5M\)
d) \(C_{M\left(CaCl_2\right)}=\dfrac{0,1}{0,4}=0,25M\)
\(n_{Mg}=\dfrac{4,8}{24}=0,2\left(mol\right)\\ Mg+2HCl\rightarrow MgCl_2+H_2\\ 0,2..............0,4.............0,2...............0,2\left(mol\right)\\ a,V_{H_2\left(đktc\right)}=0,2.22,4=4,48\left(l\right)\\ b,m_{MgCl_2}=95.0,2=19\left(g\right)\\ c,a=C_{MddHCl}=\dfrac{0,4}{0,2}=2\left(M\right)\)
PTHH: \(Mg+2HCl\rightarrow MgCl_2+H_2\uparrow\)
\(Al_2O_3+6HCl\rightarrow2AlCl_3+3H_2O\)
Ta có: \(\left\{{}\begin{matrix}\Sigma n_{HCl}=0,4\cdot2=0,8\left(mol\right)\\n_{H_2}=\dfrac{2,24}{22,4}=0,1\left(mol\right)\end{matrix}\right.\)
\(\Rightarrow\left\{{}\begin{matrix}n_{Mg}=0,1mol\\n_{Al_2O_3}=0,1mol\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}m_{Al_2O_3}=0,1\cdot102=10,2\left(g\right)\\m_{Mg}=0,1\cdot24=2,4\left(g\right)\end{matrix}\right.\)
\(\Rightarrow m_{hh}=10,2+2,4=12,6\left(g\right)\)
Theo PTHH: \(\left\{{}\begin{matrix}n_{MgCl_2}=n_{Mg}=0,1mol\\n_{AlCl_3}=2n_{Al_2O_3}=0,2mol\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}m_{MgCl_2}=0,1\cdot95=9,5\left(g\right)\\m_{AlCl_3}=0,2\cdot133,5=26,7\left(g\right)\end{matrix}\right.\)
Mặt khác: \(\left\{{}\begin{matrix}m_{ddHCl}=400\cdot1,2=480\left(g\right)\\m_{H_2}=0,1\cdot2=0,2\left(g\right)\end{matrix}\right.\)
\(\Rightarrow m_{dd}=m_{hh}+m_{ddHCl}-m_{H_2}=492,4\left(g\right)\)
\(\Rightarrow\left\{{}\begin{matrix}C\%_{MgCl_2}=\dfrac{9,5}{492,4}\cdot100\%\approx1,93\%\\C\%_{AlCl_3}=\dfrac{26,7}{492,4}\cdot100\%\approx5,42\%\end{matrix}\right.\)
a) 2Al + 6HCl -> 2AlCl3 + 3H2
Al2O3 + 6HCl -> 2AlCl3 + 3H2O
nH2 = 0,15mol => nAl=0,1mol => mAl=2,7g; mAl2O3 = 10,2g => nAl2O3 = 0,1mol
=>%mAl=20,93% =>%mAl2O3 = 79,07%
b) nHCl = 0,1.3+0,1.6=0,9 mol=>mHCl(dd)=100g
mddY=12,9+100-0,15.2=112,6g
mAlCl3=22,5g=>C%=19,98%
a)
$M + 2HCl \to MCl_2 + H_2$
$n_{HCl} = 0,3.1 = 0,3(mol)$
Theo PTHH : $n_M = \dfrac{1}{2}n_{HCl} = 0,15(mol)$
$\Rightarrow M = \dfrac{3,6}{0,15} = 24(Mg)$
b)
$n_{MgCl_2} = n_{Mg} = 0,15(mol)$
$m_{MgCl_2} = 0,15.95 = 14,25(gam)$
c) $n_{H_2} = n_{Mg} = 0,15(mol)$
$V_{H_2} = 0,15.22,4 = 3,36(lít)$
1)
a)
$CaO + 2HCl \to CaCl_2 + H_2O$
$CaCO_3 + 2HCl \to CaCl_2 + CO_2 + H_2O$
$n_{CaCO_3} = n_{CO_2} = 0,2(mol)$
$n_{CaCl_2} = 0,3(mol)$
Suy ra:
$n_{CaO} = 0,3 - 0,2 = 0,1(mol)$
$\%m_{CaO} = \dfrac{0,1.56}{0,1.56 + 0,2.100}.100\% = 21,875\%$
$\%m_{CaCO_3} = 78,125\%$
b)
$m_{dd} = 0,1.56 + 0,2.100 + 50 - 0,2.44 = 66,8(gam)$
$C\%_{CaCl_2} = \dfrac{33,3}{66,8}.100\% = 49,85\%$
Câu 4 :
a)
Gọi $n_{Fe} = a(mol) ; n_{MgO} = b(mol)$
Suy ra: $56a + 40b = 19,2(1)$
$Fe + 2HCl \to FeCl_2 + H_2$
$MgO + 2HCl \to MgCl_2 + H_2O$
Theo PTHH : $n_{HCl} = 2a + 2b = 0,4.2 = 0,8(2)$
Từ (1)(2) suy ra a = b = 0,2
$\%m_{Fe} = \dfrac{0,2.56}{19,2}.100\% = 58,33\%$
$\%m_{MgO} = 100\% -58,33\% = 41,67\%$
b)
$n_{FeCl_2} = a = 0,2(mol)$
$n_{MgCl_2} = b = 0,2(mol)$
$m_{muối} = 0,2.127 + 0,2.95 = 44,4(gam)$
a) Na2SO3 + 2HCl --> 2NaCl + SO2 + H2O
b) nHCl = 0,2.1 = 0,2 (mol)
Na2SO3 + 2HCl --> 2NaCl + SO2 + H2O
_0,1<------0,2------->0,2----->0,1
mNaCl = 0,2.58,5 = 11,7(g)
VSO2 = 0,1.22,4 = 2,24 (l)
c) mNa2SO3 = 0,1.126 = 12,6 (g)
d) \(C_{M\left(NaCl\right)}=\dfrac{0,2}{0,2}=1M\)