Cho 25,2g sắt phản ứng hoàn toàn với 200ml dd HCL 𝐚) Tính thể tích khí thoát ra (đktc) 𝐛) Tính nồng độ mol dd HCL đã dùng 𝐜) Tính nồng độ mol dd thu được sau phản ứng.
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a, Ta có: \(n_{KOH}=0,15.2=0,3\left(mol\right)\)
PT: \(KOH+HCl\rightarrow KCl+H_2O\)
Theo PT: \(n_{HCl\left(pư\right)}=n_{KOH}=0,3\left(mol\right)\)
Mà: HCl dư 15%
\(\Rightarrow n_{HCl}=0,3+0,3.15\%=0,345\left(mol\right)\)
\(\Rightarrow V_{ddHCl}=\dfrac{0,345}{1,5}=0,23\left(l\right)\)
b, Theo PT: \(n_{KCl}=n_{KOH}=0,3\left(mol\right)\)
\(\Rightarrow C_{M_{KCl}}=\dfrac{0,3}{0,15+0,23}\approx0,789\left(M\right)\)
\(C_{M_{HCl\left(dư\right)}}=\dfrac{0,3.15\%}{0,15+0,23}\approx0,118\left(M\right)\)
TTĐ:
\(m_{Fe_3O_4}=46,4\left(g\right)\)
\(C_{M_{H_2SO_4}}=2\left(M\right)\)
___________
a) \(V_{H_2SO_4}=?\left(l\right)\)
b)\(C_{M_{Fe_2\left(SO_4\right)_3}}=?\left(M\right)\)
Giải
\(n_{Fe_3O_4}=\dfrac{m}{M}=\dfrac{46,4}{232}=0,2\left(mol\right)\)
\(Fe_3O_4+4H_2SO_4\rightarrow Fe_2\left(SO_4\right)_3+FeSO_4+4H_2O\)
\(0,2\rightarrow0,8\) : 0,2 : 0,2 (mol)
\(a)V_{H_2SO_4}=\dfrac{n}{C_M}=\dfrac{0,8}{2}=0,4\left(l\right)\)
\(b)C_{M_{FeSO_{\text{4 }}}}=C_{M_{Fe_2\left(SO_4\right)_3}}=\dfrac{n}{V}=\dfrac{0,2}{0,4}=0,5\left(M\right)\)
\(a)n_{Fe_3O_4}=\dfrac{46,4}{232}=0,2mol\\Fe_3O_4+4H_2SO_4\rightarrow FeSO_4+Fe_2\left(SO_4\right)_3+4H_2O\)
0,2 0,8 0,2 0,2 0,8
\(V_{H_2SO_4}=\dfrac{0,8}{2}=0,4l\\ b)C_{M\left(FeSO_4\right)}=\dfrac{0,2}{0,4}=0,4M\\ C_{M\left(Fe_2\left(SO_4\right)_3\right)}=\dfrac{0,2}{0,4}=0,5M\)
a) \(n_{Fe}=\dfrac{1,12}{56}=0,02\left(mol\right)\)
PTHH: Fe + 2HCl -->FeCl2 + H2
_____0,02->0,04--->0,02--->0,02
=> VH2 = 0,02.22,4 = 0,448(l)
b) mFeCl2 = 0,02.127 = 2,54(g)
c) \(C_{M\left(HCl\right)}=\dfrac{0,04}{0,2}=0,2M\)
Fe + 2HCl → FeCl2 + H2
1 2 1 1
0,02 0,04 0,02 0,02
nFe=\(\dfrac{1,12}{56}\)= 0,02(mol)
a). nH2=\(\dfrac{0,02.1}{1}\)= 0,02(mol)
→VH2= n . 22,4 = 0,02 . 22,4 = 0,448(l)
b). nFeCl2= \(\dfrac{0,02.1}{1}\)= 0,02(mol)
→mFeCl2= n . M = 0,02 . 127 = 2,54(g)
c). 200ml = 0,2l
nHCl= \(\dfrac{0,02.2}{1}\)=0,04(mol)
→CM= \(\dfrac{n}{V}\)= \(\dfrac{0,04}{0,2}\)= 0,2M
PTHH: \(Fe+2HCl\rightarrow FeCl_2+H_2\uparrow\)
Ta có: \(n_{Fe}=\dfrac{5,6}{56}=0,1\left(mol\right)\)
\(\Rightarrow\left\{{}\begin{matrix}n_{HCl}=0,2\left(mol\right)\\n_{FeCl_2}=0,1\left(mol\right)=n_{H_2}\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}V_{H_2}=0,1\cdot22,4=2,24\left(l\right)\\m_{FeCl_2}=0,1\cdot127=12,7\left(g\right)\\C_{M_{FeCl_2}}=\dfrac{0,1}{0,1}=1\left(M\right)\\C_{M_{HCl}}=\dfrac{0,2}{0,1}=2\left(M\right)\end{matrix}\right.\)
\(a)n_{H_2}=\dfrac{8,96}{22,4}=0,4mol\\ n_{Fe}=a;n_{Al}=b\)
\(Fe+2HCl\rightarrow FeCl_2+H_2\)
\(2Al+6HCl\rightarrow2AlCl_3+3H_2\)
\(\Rightarrow\left\{{}\begin{matrix}56a+27b=11\\a+1,5b=0,4\end{matrix}\right.\\ \Rightarrow a=0,1;b=0,2\)
\(\%m_{Fe}=\dfrac{0,1.56}{11}\cdot100=50,91\%\\ \%m_{Al}=100-50,91=49,09\%\)
\(b)Fe+2HCl\rightarrow FeCl_2+H_2\)
0,1 0,2 0,1 0,1
\(2Al+6HCl\rightarrow2AlCl_2+3H_2\)
0,2 0,6 0,2 0,3
\(m_{HCl}=\dfrac{\left(0,2+0,6\right).36,5}{9,125}\cdot100=320g\)
\(c)m_{dd}=320+11-0,1.2-0,3.2=308,2g\)
\(C_{\%FeCl_2}=\dfrac{0,1.127}{308,2}\cdot100=4,12\%\\ C_{\%AlCl_3}=\dfrac{0,2.133,5}{308,2}\cdot100=8,66\%\)
\(PTHH:4Al+6HCl\rightarrow2Al_2Cl_3+3H_2\uparrow\)
\(n_{Al}=\frac{3,78}{27}=0,14\left(mol\right)\)
\(\Rightarrow n_{H_2}=\frac{3}{4}n_{Al}=0,105\left(mol\right)\)
\(V_{H_2}=0,105.22,4=2,352\left(l\right)\)
\(n_{HCl}=\frac{3}{2}n_{Al}=\frac{3}{2}.0,14=0,21\left(mol\right)\)
\(C_{M_{ddHCl}}=\frac{0,21}{0,2}=1,05\left(M\right)\)
\(n_{Al_2Cl_3}=\frac{1}{2}n_{Al}=\frac{1}{2}.0,14=0,07\left(mol\right)\)
\(m_{Al_2Cl_3}=0,07.160,5=11,235\left(g\right)\)
\(a)n_{Fe}=\dfrac{25,2}{56}=0,45mol\\ Fe+2HCl\rightarrow FeCl_2+H_2\)
0,45 0,9 0,45 0,45
\(V_{H_2\left(đktc\right)}=0,45.22,4=10,08l\\ b)C_{M\left(HCl\right)}=\dfrac{0,9}{0,2}=4,5M\\ c)C_{M\left(FeCl_2\right)}=\dfrac{0,45}{0,2}=2,25M\)