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a)
Gọi số mol Fe, Al là a, b (mol)
=> 56a + 27b = 19,3 (1)
\(n_{H_2}=\dfrac{14,56}{22,4}=0,65\left(mol\right)\)
PTHH: Fe + H2SO4 --> FeSO4 + H2
a--->a---------------->a
2Al + 3H2SO4 --> Al2(SO4)3 + 3H2
b---->1,5b------------------->1,5b
=> a + 1,5b = 0,65 (2)
(1)(2) => a = 0,2 (mol); b = 0,3 (mol)
mFe = 0,2.56 = 11,2 (g); mAl = 0,3.27 = 8,1 (g)
b)
\(n_{H_2SO_4}=0,65\left(mol\right)\)
=> \(C_{M\left(dd.H_2SO_4\right)}=\dfrac{0,65}{0,2}=3,25M\)
$a)PTHH:2Al+6HCl\to 2AlCl_3+3H_2$
$n_{H_2}=\dfrac{5,04}{22,4}=0,225(mol)$
$\Rightarrow n_{Al}=0,15(mol)$
$\Rightarrow \%m_{Al}=\dfrac{0,15.27}{9,45}.100\%\approx 42,86\%$
$\Rightarrow \%m_{Cu}=100-42,86=57,14\%$
$b)$ Theo PT: $n_{HCl}=2n_{H_2}=0,45(mol)$
$\Rightarrow C_{M_{HCl}}=\dfrac{0,45.110\%}{0,5}=0,99M$
\(n_{HCl}=0,3.2=0,6\left(mol\right)\\ n_{H_2}=\dfrac{5,6}{22,4}=0,25\left(mol\right)\\ 2Al+6HCl\rightarrow2AlCl_3+3H_2\\ Fe+2HCl\rightarrow FeCl_2+H_2\\ Vì:\dfrac{0,6}{2}>\dfrac{0,25}{1}\Rightarrow HCldư\\ Đặt:n_{Al}=t\left(mol\right);n_{Fe}=r\left(mol\right)\\ \left(t,r>0\right)\\ \Rightarrow\left\{{}\begin{matrix}27t+56r=8,3\\1,5t+r=0,25\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}t=0,1\\r=0,1\end{matrix}\right.\\ \Rightarrow m_{Al}=0,1.27=2,7\left(g\right);m_{Fe}=0,1.56=5,6\left(g\right)\\ b,n_{AlCl_3}=n_{Al}=0,1\left(mol\right)\Rightarrow m_{AlCl_3}=0,1.133,5=13,35\left(g\right)\\ n_{Fe}=n_{FeCl_2}=0,1\left(mol\right)\Rightarrow m_{ddFeCl_2}=127.0,1=12,7\left(g\right)\\ m_{ddHCl}=300.1,15=345\left(g\right)\\ m_{ddsau}=8,3+345-0,25.2=352,8\left(g\right)\)
\(n_{HCl\left(dư\right)}=0,6-0,25.2=0,1\left(mol\right)\\ \Rightarrow m_{ddHCl}=0,1.36,5=3,65\left(g\right)\\ C\%_{ddHCl\left(dư\right)}=\dfrac{3,65}{352,8}.100\approx1,035\%\\ C\%_{ddAlCl_3}=\dfrac{13,35}{352,8}.100\approx3,784\%\\ C\%_{ddFeCl_2}=\dfrac{12,7}{352,8}.100\approx3,6\%\)
Đặt \(\left\{{}\begin{matrix}n_{Mg}=n_{MgCl_2}=a\left(mol\right)\\n_{Fe}=n_{FeCl_2}=b\left(mol\right)\end{matrix}\right.\) \(\Rightarrow24a+56b=5,12\) (1)
Ta có: \(n_{H_2}=\dfrac{2,688}{22,4}=0,12\left(mol\right)\)
Bảo toàn electron: \(2a+2b=0,24\) (2)
Từ (1) và (2) \(\Rightarrow\left\{{}\begin{matrix}a=n_{MgCl_2}=0,05\left(mol\right)\\b=n_{FeCl_2}=0,07\left(mol\right)\end{matrix}\right.\)
Bảo toàn nguyên tố: \(n_{HCl\left(p/ứ\right)}=2n_{MgCl_2}+2n_{FeCl_2}=0,24\left(mol\right)\)
PTHH: \(NaOH+HCl\rightarrow NaCl+H_2O\)
Theo PTHH: \(n_{HCl\left(dư\right)}=n_{NaOH}=0,06\left(mol\right)\)
\(\Rightarrow\Sigma n_{HCl}=0,3\left(mol\right)\) \(\Rightarrow m_{ddHCl}=\dfrac{0,3\cdot36,5}{36,5\%}=30\left(g\right)\)
Mặt khác: \(m_{H_2}=0,12\cdot2=0,24\left(g\right)\)
\(\Rightarrow m_{dd}=m_{KL}+m_{ddHCl}-m_{H_2}=34,88\left(g\right)\)
\(\Rightarrow\left\{{}\begin{matrix}C\%_{FeCl_2}=\dfrac{0,07\cdot127}{34,88}\cdot100\%\approx25,49\%\\C\%_{MgCl_2}=\dfrac{0,05\cdot95}{34,88}\cdot100\%\approx13,62\%\\C\%_{HCl\left(dư\right)}=\dfrac{0,04\cdot36,5}{34,88}\cdot100\%\approx4,19\%\end{matrix}\right.\)
\(n_{Fe}=a\left(mol\right),n_{Mg}=b\left(mol\right)\)
\(m_{hh}=56a+24b=10.16\left(g\right)\)
\(n_{H_2}=\dfrac{5.6}{22.4}=0.25\left(mol\right)\)
\(Fe+2HCl\rightarrow FeCl_2+H_2\)
\(Mg+2HCl\rightarrow MgCl_2+H_2\)
\(n_{H_2}=a+b=0.25\left(mol\right)\left(2\right)\)
\(\left(1\right),\left(2\right):a=0.13,b=0.12\)
\(m_{Fe}=0.13\cdot56=7.28\left(g\right)\)
\(m_{Mg}=0.12\cdot24=2.88\left(g\right)\)
\(n_{HCl}=2\cdot n_{H_2}=2\cdot0.25=0.5\left(mol\right)\)
\(C_{M_{HCl}}=\dfrac{0.5}{0.5}=1\left(M\right)\)
a) NaOH + HCl → NaCl + H2O
Fe(OH)3 + 3HCl → FeCl3 + 3H2O
Gọi \(n_{NaOH}=x\left(mol\right);n_{Fe\left(OH\right)_3}=y\left(mol\right)\)
=> 40x+107y=29,4
n HCl = x + 3y = 0,2.4=0,8
=> x=0,2 ; y=0,2
=> % NaOH= 27,21% ; %Fe(OH)3=72,79%
b) \(n_{NaCl}=0,2\left(mol\right);n_{FeCl_3}=0,2\left(mol\right)\)
=> \(CM_{NaCl}=\dfrac{0,2}{0,2}=1M\)
\(CM_{FeCl_3}=\dfrac{0,2}{0,2}=1M\)