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1.
\(m_{HCl}=\dfrac{10,95.75}{100}=8,2125\left(g\right)\)
\(\Rightarrow n_{HCl}=\dfrac{8,2125}{35,5}=0,225\left(mol\right)\)
\(Fe_2O_3+6HCl\rightarrow2FeCl_3+3H_2O\)
\(\Rightarrow n_{Fe_2O_3}=\dfrac{1}{6}n_{HCl}=0,0375\left(mol\right)\)
\(n_{FeCl_3}=\dfrac{1}{3}n_{HCl}=0,075\left(mol\right)\)
\(\Rightarrow C\%=\dfrac{0,075.162,5}{0,0375.160+75}.100\%=15,05\%\)
\(a,2NaOH+MgSO_4\rightarrow Mg\left(OH\right)_2+Na_2SO_4\\ n_{NaOH}=0,5.1=0,5\left(mol\right)\\ b,n_{Mg\left(OH\right)_2}=\dfrac{0,5}{2}=0,25\left(mol\right)=n_{Na_2SO_4}\\ m_{kt}=m_{Mg\left(OH\right)_2}=58.0,25=14,5\left(g\right)\\ c,V_{ddX}=V_{ddNaOH}+V_{ddMgSO_4}=0,5+0,5=1\left(l\right)\\ C_{MddNa_2SO_4}=\dfrac{0,25}{1}=0,25\left(M\right)\)
Gọi: \(\left\{{}\begin{matrix}n_{FeCl_3}=x\left(mol\right)\\n_{MgCl_2}=y\left(mol\right)\end{matrix}\right.\)
PT: \(FeCl_3+3KOH\rightarrow Fe\left(OH\right)_{3\downarrow}+3KCl\)
______x_________3x_________x (mol)
\(MgCl_2+2KOH\rightarrow Mg\left(OH\right)_{2\downarrow}+2KCl\)
____y_________2y_________y (mol)
Ta có: \(n_{KOH}=0,2.2,5=0,5\left(mol\right)\)
⇒ 3x + 2y = 0,5 (1)
m kết tủa = 16,5 ⇒ 107x + 58y = 16,5 (2)
Từ (1) và (2) ⇒ x = y = 0,1 (mol)
\(\Rightarrow C_{M_{FeCl_3}}=C_{M_{MgCl_2}}=\dfrac{0,1}{0,5}=0,2\left(M\right)\)
\(FeCl_3+3KOH\rightarrow Fe\left(OH\right)_3\downarrow+3KCl\\ MgCl_2+2KOH\rightarrow Mg\left(OH\right)_2\downarrow+2KCl\)
\(n_{KOH}=0,2\cdot2,5=0,5\left(mol\right)\)
Đặt nFeCl₃ trong 500ml X là a mol, nMgCl₂ trong 500ml X là b mol
\(\Rightarrow\left\{{}\begin{matrix}3a+2b=0,5\\107a+58b=16,5\end{matrix}\right.\\ \Leftrightarrow\left\{{}\begin{matrix}a=0,1\\b=0,1\end{matrix}\right.\)
\(\Rightarrow C_MFeCl_3=\dfrac{0,1}{0,5}=0,2\left(M\right)\\ C_MMgCl_2=\dfrac{0,1}{0,5}=0,2\left(M\right)\)
\(n_{CuSO_4}=\dfrac{15,2}{160}=0,095mol\\ CuSO_4+2NaOH\rightarrow Cu\left(OH\right)_2+Na_2SO_4\)
0,095 0,19 0,095 0,095
\(m_{rắn}=m_{Cu\left(OH\right)_2}=0,095.98=9,31g\\ V_{ddNaOH}=\dfrac{0,19}{2}=0,095l\\ b)C_{M_{Na_2SO_4}}=\dfrac{0,095}{0,04+0,095}\approx0,7M\\ c)Cu\left(OH\right)_2\xrightarrow[t^0]{}CuO+H_2O\)
0,095 0,095
\(m_{rắn}=m_{CuO}=0,095.80=7,6g\)
\(a,n_{HCl}=3.0,2=0,6(mol)\\ PTHH:X(OH)_n+nHCl\to XCl_n+nH_2O\\ \Rightarrow n.n_{X(OH)_n}=n_{HCl}=0,6(mol)\\ \Rightarrow M_{X(OH)_n}=\dfrac{15,6n}{0,6}=26n\\ \Rightarrow M_X+17n=26n\\ \Rightarrow M_X=9n\)
Thay \(m=3\Rightarrow M_X=27(g/mol)\)
Vậy X là nhôm (Al) và CT của bazơ là \(Al(OH)_3\)
\(b,n_{Al(OH)_3}=\dfrac{15,6}{78}=0,2(mol)\\ n_{H_2SO_4}=\dfrac{196.20\%}{100\%.98}=0,4(mol)\\ PTHH:2Al(OH)_3+3H_2SO_4\to Al_2(SO_4)_3+6H_2O\)
Vì \(\dfrac{n_{Al(OH)_3}}{2}<\dfrac{n_{H_2SO_4}}{3}\) nên \(H_2SO_4\) dư
\(\Rightarrow n_{Al_2(SO_4)_3}=\dfrac{1}{2}n_{Al(OH)_3}=0,1(mol)\\ \Rightarrow C\%_{Al_2(SO_4)_3}=\dfrac{0,1.342}{15,6+196}.100\%=16,16\%\)
Ta có: \(n_{H_2}=\dfrac{4,48}{22,4}=0,2\left(mol\right)\)
a. PTHH: \(Mg+2HCl--->MgCl_2+H_2\)
Theo PT: \(n_{Mg}=n_{H_2}=0,2\left(mol\right)\)
=> \(m_{Mg}=0,2.24=4,8\left(g\right)\)
Theo PT: \(n_{HCl}=2.n_{Mg}=2.0,2=0,4\left(mol\right)\)
=> \(m_{HCl}=0,4.36.5=14,6\left(g\right)\)
=> \(C_{\%_{HCl}}=\dfrac{14,6}{200}.100\%=7,3\%\)
b. Ta có: \(m_{dd_{MgCl_2}}=4,8+200=204,8\left(g\right)\)
Theo PT: \(n_{MgCl_2}=n_{Mg}=0,2\left(mol\right)\)
=> \(m_{MgCl_2}=0,2.95=19\left(g\right)\)
=> \(C_{\%_{MgCl_2}}=\dfrac{19}{204,8}.100\%=9,28\%\)