hoà tan hoàn toàn kim loại Zn trong 200g dung dịch HCL sau phản ứng thu được 7,84 lít khí ĐKTC
a) tính khối lượng Zn đã dùng
b) tính nồng độ % của dung dịch HCL
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\(n_{H_2}=\dfrac{2,24}{22,4}=0,1=n_{Zn}\\ n_{ZnO}=\dfrac{14,6-6,5}{81}=0,1mol\\ C\%=\dfrac{0,2\cdot136}{175,6+14,6-0,2}=14,32\%\)
\(n_{Zn} = a(mol) ; n_{Al} = b(mol) ; n_{Mg} = c(mol)\\ \Rightarrow 65a + 27b + 24c = 44,1(1)\\ Zn + 2HCl \to ZnCl_2 + H_2\\ 2Al + 6HCl \to 2AlCl_3 + 3 H_2\\ Mg + 2HCl \to MgCl_2 + H_2\\ n_{H_2} = a + 1,5b + c = \dfrac{31,36}{22,4} = 1,4(2)\\ Mà : 2a = 3b(3)\\ (1)(2)(3) \Rightarrow a = 0,3 ; b = 0,2 ; c = 0,8\\ \%m_{Zn} = \dfrac{0,3.65}{44,1}.100\% = 44,22\%\\ \%m_{Al} = \dfrac{0,2.27}{44,1}.100\% = 12,24\%\)
\(\%m_{Mg} = 100\% -44,22\% -12,24\% = 43,54\%\)
Gọi x, y lần lượt là số mol của Zn và Fe
Ta có: \(n_{H_2}=\dfrac{6,72}{22,4}=0,3\left(mol\right)\)
a. PTHH:
Zn + 2HCl ---> ZnCl2 + H2 (1)
Fe + 2HCl ---> FeCl2 + H2 (2)
Theo PT(1): \(n_{H_2}=n_{Zn}=x\left(mol\right)\)
Theo PT(2): \(n_{H_2}=n_{Fe}=y\left(mol\right)\)
=> x + y = 0,3 (*)
Theo đề, ta có: 65x + 56y = 17,7 (**)
Từ (*) và (**), ta có HPT:
\(\left\{{}\begin{matrix}x+y=0,3\\65x+56y=17,7\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=0,1\\y=0,2\end{matrix}\right.\)
=> \(m_{Zn}=0,1.65=6,5\left(g\right)\)
=> \(\%_{m_{Zn}}=\dfrac{6,5}{17,7}.100\%=36,72\%\)
\(\%_{m_{Fe}}=100\%-36,72\%=63,28\%\)
b. Ta có: \(n_{hh_{Zn,Fe}}=0,1+0,2=0,3\left(mol\right)\)
Theo PT(1, 2): \(n_{HCl}=2.n_{hh}=2.0,3=0,6\left(mol\right)\)
=> \(m_{HCl}=0,6.36,5=21,9\left(g\right)\)
=> \(C_{\%_{HCl}}=\dfrac{21,9}{200}.100\%=10,95\%\)
a)
\(n_{Zn}=\dfrac{13}{65}=0,2\left(mol\right)\)
PTHH: Zn + 2HCl --> ZnCl2 + H2
0,2-->0,4----->0,2--->0,2
=> VH2 = 0,2.22,4 = 4,48 (l)
b) mHCl = 0,4.36,5 = 14,6 (g)
=> \(m_{dd.HCl}=\dfrac{14,6.100}{7,3}=200\left(g\right)\)
c)
mdd sau pư = 13 + 200 - 0,2.2 = 212,6 (g)
mZnCl2 = 0,2.136 = 27,2 (g)
=> \(C\%=\dfrac{27,2}{212,6}.100\%=12,8\%\)
\(a.Zn+2HCl\rightarrow ZnCl_2+H_2\\ n_{Zn}=n_{ZnCl_2}=n_{H_2}=0,25\left(mol\right)\\ \Rightarrow m_{Zn}=0,25.65=16,25\left(g\right)\\ m_{ZnCl_2}=0,25.136=34\left(g\right)\\ b.FeO+H_2-^{t^o}\rightarrow Fe+H_2O\\ Tacó:n_{Fe}=n_{H_2}=0,25\left(mol\right)\\ \Rightarrow m_{Fe}=0,25.56=14\left(g\right)\)
a)
Gọi $n_{Zn} = a(mol) ; n_{Al} = b(mol) \Rightarrow 65a + 27b = 11,9(1)$
$Zn + 2HCl \to ZnCl_2 + H_2$
$2Al + 6HCl \to 2AlCl_3 + 3H_2$
Theo PTHH :
$n_{H_2} = a + 1,5b = \dfrac{8,96}{22,4} = 0,4(2)$
Từ (1)(2) suy ra : a = 0,1; b = 0,2
$m_{Zn} = 0,1.65 = 6,5(gam)$
$m_{Al} = 0,2.27 = 5,4(gam)$
b) $n_{HCl} = 2n_{H_2} = 0,8(mol)$
$C\%_{HCl} = \dfrac{0,8.36,5}{125}.100\% = 23,36\%$
\(n_{HCl}=\dfrac{200.14,6\%}{36,5}=0,8\left(mol\right)\\ Đặt:\left\{{}\begin{matrix}n_{Zn}=x\left(mol\right)\\n_{Mg}=y\left(mol\right)\end{matrix}\right.\\ Zn+2HCl\rightarrow ZnCl_2+H_2\\ Mg+2HCl\rightarrow MgCl_2+H_2\\ Tacó:\left\{{}\begin{matrix}65x+24y=12,5\\x+y=0,35\left(mol\right)\end{matrix}\right.\\ \Rightarrow\left\{{}\begin{matrix}x=0,1\\y=0,25\end{matrix}\right.\\ \Rightarrow m_{Zn}=6,5\left(g\right);m_{Mg}=6\left(g\right)\\ b.Tacó:BTNT\left(H\right):n_{HCl}.1>n_{H_2}.2\\ \Rightarrow HCldưsauphảnứng\\ Dungdịchsauphảnứnggồm:\left\{{}\begin{matrix}ZnCl_2:0,1\left(mol\right)\\MgCl_2:0,25\left(mol\right)\\HCl_{dư}:0,8-0,7=0,1\left(mol\right)\end{matrix}\right.\\ m_{ddsaupu}=200+12,5-0,35.2=212,8\left(g\right)\\ C\%_{ZnCl_2}=\dfrac{0,1.136}{212,8}.100=6,39\%;C\%_{MgCl_2}=\dfrac{0,25.95}{212,8}.100=11,16\%;C\%_{HCl\left(dư\right)}=\dfrac{0,1.36,5}{212,8}.100=1,72\%\)
C32:
a, \(n_C=\dfrac{2,4}{12}=0,2\left(mol\right)\)
PT: \(C+O_2\underrightarrow{t^o}CO_2\)
Theo PT: \(n_{CO_2}=n_C=0,2\left(mol\right)\Rightarrow V_{CO_2}=0,2.22,4=4,48\left(l\right)\)
b, \(n_{NaOH}=0,3.1=0,3\left(mol\right)\)
\(\Rightarrow\dfrac{n_{NaOH}}{n_{CO_2}}=1,5\) → Pư tạo NaHCO3 và Na2CO3
PT: \(CO_2+NaOH\rightarrow NaHCO_3\)
\(CO_2+2NaOH\rightarrow Na_2CO_3+H_2O\)
Ta có: \(\left\{{}\begin{matrix}n_{CO_2}=n_{NaHCO_3}+n_{Na_2CO_3}=0,2\\n_{NaOH}=n_{NaHCO_3}+2n_{Na_2CO_3}=0,3\end{matrix}\right.\)
\(\Rightarrow\left\{{}\begin{matrix}n_{NaHCO_3}=0,1\left(mol\right)\\n_{Na_2CO_3}=0,1\left(mol\right)\end{matrix}\right.\)
⇒ mNaHCO3 = 0,1.84 = 8,4 (g)
mNa2CO3 = 0,1.106 = 10,6 (g)
c, \(C_{M_{NaHCO_3}}=C_{M_{Na_2CO_3}}=\dfrac{0,1}{0,3}=\dfrac{1}{3}\left(M\right)\)
Lần sau bạn đăng tách câu hỏi ra nhé.
C31:
a, \(Zn+2HCl\rightarrow ZnCl_2+H_2\)
\(ZnO+2HCl\rightarrow ZnCl_2+H_2O\)
b, \(n_{H_2}=\dfrac{2,24}{22,4}=0,1\left(mol\right)\)
Theo PT: \(n_{Zn}=n_{H_2}=0,1\left(mol\right)\)
\(\Rightarrow\left\{{}\begin{matrix}\%m_{Zn}=\dfrac{0,1.65}{14,6}.100\%\approx44,52\%\\\%m_{ZnO}\approx55,48\%\end{matrix}\right.\)
c, \(n_{ZnO}=\dfrac{14,6-0,1.65}{81}=0,1\left(mol\right)\)
Theo PT: \(n_{HCl}=2n_{Zn}+2n_{ZnO}=0,4\left(mol\right)\)
\(\Rightarrow m_{ddHCl}=\dfrac{0,4.36,5}{10\%}=146\left(g\right)\)
\(n_{H_2}=\dfrac{7,84}{22,4}=0,35\left(mol\right)\)
PT: \(Zn+2HCl\rightarrow ZnCl_2+H_2\)
____0,35_____0,7___________0,35 (mol)
a, \(m_{Zn}=0,35.65=22,75\left(g\right)\)
b, \(C\%_{HCl}=\dfrac{0,7.36,5}{200}.100\%=12,775\%\)
\(a)n_{H_2}=\dfrac{7,84}{22,4}=0,35mol\\ Zn+2HCl\rightarrow ZnCl_2+H_2\)
0,35 0,7 0,35 0,35
\(m_{Zn}=0,35.65=22,75g\\ b)C_{\%HCl}=\dfrac{0,7.36,5}{200}\cdot100\%=12,775\%\)