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\(n_{Fe}=\dfrac{36,4}{56}=0,65\left(mol\right)\)
PTHH: Fe + 2HCl ---> FeCl2 + H2
0,65->1,3----->0,65--->0,65
=> \(\left\{{}\begin{matrix}a,V_{ddHCl}=\dfrac{1,3}{0,5}=2,6\left(l\right)\\b,V_{H_2}=0,65.22,4=14,56\left(l\right)\end{matrix}\right.\)
c, \(C_{M\left(FeCl_2\right)}=\dfrac{0,65}{2,6}=0,25M\)
\(a,n_{Zn}=\dfrac{8,125}{65}=0,125\left(mol\right)\)
PTHH: Zn + 2HCl ---> ZnCl2 + H2
0,125->0,25----->0,125->0,125
\(\Rightarrow\left\{{}\begin{matrix}a,V_{ddHCl}=\dfrac{0,25}{0,5}=0,5\left(l\right)\\b,V_{H_2}=0,125.22,4=2,8\left(l\right)\\c,C_{M\left(ZnCl_2\right)}=\dfrac{0,125}{0,5}=0,25M\end{matrix}\right.\)
Ta có: nHCl(trong V1) = 0,5V1 (mol)
nHCl(trong V2 ) = 3V2 (mol)
nHCl(sau khi trộn) = 0,1 x 2,5 = 0,25 (mol)
=> 0,5V1 + 3V2 = 0,25 (1)
Lại có: Thể tích dung dịch thu được là 100(ml) = 0,1 (lít)
=> V1 + V2 = 0,1 (2)
Từ (1), (2), ta có hệ phương trình:
\(\left\{\begin{matrix}0,5V_1+3V_2=0,25\\V_1+V_2=0,1\end{matrix}\right.\)
\(\Rightarrow\left\{\begin{matrix}V_1=0,02\left(l\right)=20\left(ml\right)\\V_2=0,08\left(l\right)=80\left(ml\right)\end{matrix}\right.\)
\(n_{H_2}=\dfrac{11.2}{22.4}=0.5\left(mol\right)\)
\(a.\)
\(n_{HCl}=2n_{H_2}=0.5\cdot2=1\left(mol\right)\)
\(V_{dd_{HCl}}=\dfrac{1}{2}=0.5\left(M\right)\)
\(b.\)
\(n_X=a\left(mol\right)\)
\(\Rightarrow n_Y=2a\left(mol\right),n_Z=a\left(mol\right),n_T=a\left(mol\right)\)
\(M_X=M\left(\text{g/mol}\right)\)
\(\Rightarrow M_Y=2.7M\left(\text{g/mol}\right),M_Z=\dfrac{7M}{3}\left(\text{g/mol}\right),M_T=\dfrac{347}{60}M\left(\text{g/mol}\right)\)
\(m_{hh}=aM+2a\cdot2.7M+a\cdot\dfrac{7}{3}M+a\cdot\dfrac{347}{60}M=34.7\left(g\right)\)
\(\Rightarrow aM=2.4\)
\(n_{hh}=n_{H_2}=0.5\left(mol\right)\)
\(\Rightarrow a+2a+a+a=0.5\)
\(\Rightarrow a=0.1\)
\(M=\dfrac{2.4}{0.1}=24\left(\text{g/mol}\right)\Rightarrow Mg\)
\(Y=2.7\cdot24=65\left(\text{g/mol}\right)\Rightarrow Zn\)
\(Z=\dfrac{7}{3}\cdot24=56\left(\text{g/mol}\right)\Rightarrow Fe\)
\(T=\dfrac{347}{60}\cdot24=137\left(\text{g/mol}\right)\Rightarrow Ba\)
Câu `3:`
`n_[Mg]=[2,4]/24=0,1(mol)`
`Mg + 2HCl -> MgCl_2 + H_2 \uparrow`
`0,1` `0,2` `0,1` `0,1` `(mol)`
`a)C%_[HCl]=[0,2.36,5]/200 . 100=3,65%`
`b)m_[MgCl_2]=0,1.95=9,5(g)`
`c)V_[H_2]=0,1.22,4=2,24(l)`
~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
Câu `4:`
`n_[Zn]=[3,25]/65=0,05(mol)`
`Zn + 2HCl -> ZnCl_2 + H_2 \uparrow`
`0,05` `0,1` `0,05` `0,05` `(mol)`
`a)C%_[HCl]=[0,1.36,5]/200 .100=1,825%`
`b)m_[ZnCl_2]=0,05.136=6,8(g)`
`c)V_[H_2]=0,05.22,4=1,12(l)`
\(a) Fe + 2HCl \to FeCl_2 + H_2\\ b) n_{H_2} = n_{Fe} = \dfrac{5,6}{56} = 0,1(mol)\\ V_{H_2} = 0,1.22,4 = 2,24(lít)\\ c) n_{FeCl_2} = n_{Fe} = 0,1(mol)\\ m_{FeCl_2} = 0,1.127 = 12,7(gam)\\ d) n_{HCl} = 2n_{Fe} = 0,2(mol) \Rightarrow C_{M_{HCl}} = \dfrac{0,2}{0,1} = 2M\\ e)n_{Fe_3O_4} = \dfrac{2,32}{232} = 0,01(mol)\\ Fe_3O_4 + 4H_2 \xrightarrow{t^o} 3Fe + 4H_2O\\ 4n_{Fe_3O_4} = 0,04 < n_{H_2} = 0,1 \to H_2\ dư\\ \)
\(n_{Fe} = 3n_{Fe_3O_4} = 0,03(mol)\\ m_{Fe} = 0,03.56 = 1,68(gam)\)
nAl=\(\dfrac{5,8}{27}\)≈0,215 mol
a, PTPƯ: 2Al + 6HCl ---> 2AlCl3 + 3H2
Ta có: 2 mol Al ---> 3 mol H2
nên 0,215 mol Al ---> 0,323 mol H2
=> VH2=0,323.22,4≈7,24 l
b, Ta có: 2 mol Al ---> 6 mol HCl
nên 0,215 mol Al ---> 0,65 mol HCl
=> VHCl=0,65.22,4=14,56 l
c, Ta có: 2 mol Al ---> 2 mol AlCl3
nên 0,215 mol Al ---> 0,215 mol AlCl3
=> mAlCl3=0,215.133,5≈28,7 g
PTHH: \(Zn+2HCl\rightarrow ZnCl_2+H_2\uparrow\)
Ta có: \(n_{Zn}=\dfrac{0,65}{65}=0,01\left(mol\right)\)
\(\Rightarrow\left\{{}\begin{matrix}n_{HCl}=0,02\left(mol\right)\\n_{ZnCl_2}=0,01\left(mol\right)=n_{H_2}\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}C_{M_{HCl}}=\dfrac{0,02}{0,05}=0,4\left(M\right)\\m_{ZnCl_2}=0,01\cdot136=1,36\left(g\right)\\V_{H_2}=0,01\cdot22,4=0,224\left(l\right)\end{matrix}\right.\)
\(n_{HCl} = 0,6.2 = 1,2(mol)\\ \Rightarrow V_{dd\ HCl\ 0,75M} = \dfrac{1,2}{0,75} = 1,6(lít) = 1600(ml)\\ \Rightarrow V_{nước} = 1600 -600 = 1000(ml)\)