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PTHH: \(Zn+2HCl\rightarrow ZnCl_2+H_2\uparrow\)
Ta có: \(n_{H_2}=\dfrac{6,72}{22,4}=0,3\left(mol\right)\)
\(\Rightarrow\left\{{}\begin{matrix}n_{Zn}=0,3\left(mol\right)=n_{ZnCl_2}\\n_{HCl}=0,6\left(mol\right)\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}\%m_{Zn}=\dfrac{0.3\cdot65}{35,5}\cdot100\%\approx54,93\%\\\%m_{Cu}=45,07\%\\C\%_{HCl}=\dfrac{0,6\cdot36,5}{500}\cdot100\%=4,38\%\\m_{ZnCl_2}=0,3\cdot136=40,8\left(g\right)\end{matrix}\right.\)
Mặt khác: \(\left\{{}\begin{matrix}m_{Cu}=35,5-0,3\cdot65=16\left(g\right)\\m_{H_2}=0,3\cdot2=0,6\left(g\right)\end{matrix}\right.\)
\(\Rightarrow m_{dd}=m_{KL}+m_{ddHCl}-m_{Cu}-m_{H_2}=518,9\left(g\right)\)
\(\Rightarrow C\%_{ZnCl_2}=\dfrac{40,8}{518,9}\cdot100\%\approx7,86\%\)
a)
$CaCO_3 + 2HCl \to CaCl_2 + CO_2 + H_2O$
$n_{CaCO_3} = n_{CO_2} = \dfrac{672}{1000.22,4} = 0,03(mol)$
$n_{HCl} = 2n_{CO_2} = 0,06(mol)$
$\Rightarrow C_{M_{HCl}} = \dfrac{0,06}{0,2} = 0,3M$
b)
$\%m_{CaCO_3} = \dfrac{0,03.100}{5}.100\% = 60\%$
$\%m_{CaSO_4}= 100\% -60\% = 40\%$
2Al + 6HCl -> 2AlCl3 + 3H2 (1)
ZnO + 2HCl -> ZnCl2 + H2O (2)
a) nH2= 13,44/22.4=0.6(mol) -> mH2=0,6.2=1,2(g)
Theo PTHH: nAl = 2/3 nH2 = 2/3 . 0,6= 0,4(mol) -> mAl = 0,4 . 27=10,8(g)
-> mZnO = 27-10,8= 16,2(g)
b) nZnO = 16,2/81=0,2(mol)
Theo PTHH (2): nHCl = 2nZnO=2.0,2=0,4(mol)
Theo PTHH (1) : nHCl=2nH2=2.0,6=1,2(mol)
-> \(\Sigma\)nHCl = 0,4+1,2=1,6(mol)
-> mHCl = 1,6.36,5= 58,4(g)
-> mddHCl = 58,4.100/29,2= 200(g)
c) Theo PTHH (1): nAlCl3 = 2/3 nH2 = 2/3 . 0,6=0,4(mol) -> mAlCl3=0,4.133,5=53,4(g)
mdd sau phản ứng= mA + mddHCl - mH2 =27+200-1,2 =225,8(g)
-> C% AlCl3 = 53,4.100%/225,8 = 20,88%
Theo PTHH (2) nZnCl2 =nZnO= 0,2(mol)-> mZnCl2=0,2.136=27,2(g)
-> C% ZnCl2= 27,2.100%/255,8=10,63%
200ml = 0,2l
\(n_{Ba\left(OH\right)2}=0,5.0,2=0,1\left(mol\right)\)
Pt : \(Ba\left(OH\right)_2+2HCl\rightarrow BaCl_2+2H_2O|\)
1 2 1 2
0,1 0,2 0,1
a) \(n_{HCl}=\dfrac{0,1.2}{1}=0,2\left(mol\right)\)
\(V_{ddHCl}=\dfrac{0,2}{1}=0,2\left(l\right)=200\left(ml\right)\)
b) \(n_{BaCl2}=\dfrac{0,2.1}{2}=0,1\left(mol\right)\)
⇒ \(m_{BaCl2}=0,1.208=20,8\left(g\right)\)
c) \(V_{ddspu}=0,2+0,2=0,4\left(l\right)\)
\(C_{M_{BaCl2}}=\dfrac{0,1}{0,4}=0,25\left(M\right)\)
Chúc bạn học tốt
PTHH: \(Ba\left(OH\right)_2+2HCl\rightarrow BaCl_2+2H_2O\)
Ta có: \(n_{Ba\left(OH\right)_2}=0,2\cdot0,5=0,1\left(mol\right)\)
\(\Rightarrow\left\{{}\begin{matrix}n_{HCl}=0,2\left(mol\right)\\n_{BaCl_2}=0,1\left(mol\right)\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}V_{ddHCl}=\dfrac{0,2}{1}=0,2\left(l\right)=200\left(ml\right)\\m_{BaCl_2}=0,1\cdot208=20,8\left(g\right)\\C_{M_{BaCl_2}}=\dfrac{0,1}{0,2+0,2}=0,25\left(M\right)\end{matrix}\right.\)