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\(n_{Fe_2O_3}=\dfrac{32}{160}=0,2\left(mol\right)\)
\(m_{H_2SO_4}=196.40\%=78,4\left(g\right)\Rightarrow n_{H_2SO_4}=\dfrac{78,4}{98}=0,8\left(mol\right)\)
PTHH: Fe2O3 + 3H2SO4 → Fe2(SO4)3 + 3H2O
Mol: 0,2 0,6 0,2
Ta có: \(\dfrac{0,2}{1}< \dfrac{0,8}{3}\) ⇒ Fe2O3 hết, H2SO4 dư
mdd sau pứ = 32 + 196 = 228 (g)
\(C\%_{ddFe_2\left(SO_4\right)_3}=\dfrac{0,2.400.100\%}{228}=35,09\%\)
\(C\%_{ddH_2SO_4dư}=\dfrac{\left(0,8-0,6\right).98.100\%}{228}=8,596\%\)
\(n_{HCl}=\dfrac{14,6\%.450}{36,5}=1,8\left(mol\right)\\ n_{Fe_2O_3}=\dfrac{32}{160}=0,2\left(mol\right)\\ Fe_2O_3+6HCl\rightarrow2FeCl_3+3H_2O\\ Vì:\dfrac{1,8}{6}>\dfrac{0,2}{1}\\ \Rightarrow HCldư\\ a.n_{FeCl_3}=0,2.2=0,4\left(mol\right)\\ m_{FeCl_3}=162,5.0,4=65\left(g\right)\\ b.n_{HCl\left(dư\right)}=1,8-6.0,2=0,6\left(mol\right)\\ m_{HCl\left(dư\right)}=0,6.36,5=21,9\left(g\right)\\ c.m_{ddsau}=32+450=482\left(g\right)\\ C\%_{ddHCl\left(dư\right)}=\dfrac{21,9}{482}.100\approx4,544\%\\ C\%_{ddFeCl_3}=\dfrac{65}{482}.100\approx13,485\%\)
\(n_{Fe2O3}=\dfrac{32}{160}=0,2\left(mol\right)\)
\(m_{ct}=\dfrac{14,6.450}{100}=65,7\left(g\right)\)
\(n_{HCl}=\dfrac{65,7}{36,5}=1,8\left(mol\right)\)
Pt : \(Fe_2O_3+6HCl\rightarrow2FeCl_3+3H_2O|\)
1 6 2 3
0,2 1,8 0,4
a) Lập tỉ số so sánh : \(\dfrac{0,2}{1}< \dfrac{1,8}{6}\)
⇒ Fe2O3 phản ứng hết , HCl dư
⇒ Tính toán dựa vào số mol của Fe2O3
\(n_{FeCl3}=\dfrac{0,2.2}{1}=0,4\left(mol\right)\)
⇒ \(m_{FeCl3}=0,4.162,5=65\left(g\right)\)
b) \(n_{HCl\left(dư\right)}=1,8-\left(0,2.6\right)=0,6\left(mol\right)\)
⇒ \(m_{HCl\left(dư\right)}=0,6.36,5=21,9\left(g\right)\)
c) \(m_{ddspu}=32+450=482\left(g\right)\)
\(C_{FeCl3}=\dfrac{65.100}{482}=13,48\)0/0
\(C_{HCl\left(dư\right)}=\dfrac{21,9.100}{482}=4,54\)0/0
Chúc bạn học tốt
a)\(n_{Fe_2O_3}=0,2\left(mol\right)\)
PT:\(Fe_2O_3+6HCl\rightarrow2FeCl_3+3H_2O\)
\(0,2\) \(1,2\) \(0,4\)
\(\Rightarrow n_{FeCl_3}=0,4\left(mol\right)\)
\(\Rightarrow m_{FeCl_3}=65\left(g\right)\)
b) \(n_{HCl}=\dfrac{218.30\%}{35,5+1}=\dfrac{654}{365}\left(mol\right)\)
Từ PT \(\Rightarrow\)\(n_{HClpư}=1,2\left(mol\right)\)
\(\Rightarrow n_{HCldư}=\dfrac{654}{365}-1,2=\dfrac{216}{365}\left(mol\right)\)
\(\Rightarrow m_{HCldư}=21,6\left(g\right)\)
\(m_{dd}=32+218=250\left(g\right)\)
\(C\%_{FeCl_3}=\dfrac{65}{250}.100\%=26\left(\%\right)\)
\(C\%_{HCldu}=\dfrac{21,6}{250}.100\%=8,64\%\)
PTHH: \(Fe_2O_3+6HCl\rightarrow2FeCl_3+3H_2O\)
\(2FeCl_3+Cu\rightarrow CuCl_2+2FeCl_2\)
Ta có: \(\left\{{}\begin{matrix}n_{FeCl_3}=2n_{Fe_2O_3}=2\cdot\dfrac{16}{160}=0,2\left(mol\right)\\n_{Cu}=\dfrac{32}{64}=0,5\left(mol\right)\end{matrix}\right.\)
Xét tỉ lệ: \(\dfrac{0,2}{2}< \dfrac{0,5}{1}\) \(\Rightarrow\) Cu còn dư
\(\Rightarrow\left\{{}\begin{matrix}n_{CuCl_2}=0,1\left(mol\right)\\n_{FeCl_2}=0,2\left(mol\right)\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}m_{CuCl_2}=0,1\cdot135=13,5\left(g\right)\\m_{FeCl_2}=0,2\cdot127=25,4\left(g\right)\end{matrix}\right.\)
\(a.2Al+3H_2SO_4\rightarrow Al_2\left(SO_4\right)_3+3H_2\\ Fe+H_2SO_4\rightarrow FeSO_4+H_2\\ Đặt:\left\{{}\begin{matrix}n_{Al}=x\left(mol\right)\\n_{Fe}=y\left(mol\right)\end{matrix}\right.\\ Tacó:\left\{{}\begin{matrix}27x+56y=17,6\\1,5x+y=\dfrac{61}{112}\end{matrix}\right.\\ \Rightarrow\left\{{}\begin{matrix}x=\dfrac{43}{190}\\y=\dfrac{2183}{10640}\end{matrix}\right.\\ \Rightarrow\left\{{}\begin{matrix}\%m_{Al}=34,72\%\\m_{Fe}=65,28\%\end{matrix}\right.\\ b.BTNT\left(H\right):n_{H_2SO_4}=n_{H_2}=\dfrac{61}{112}\left(mol\right)\\ \Rightarrow m_{ddH_2SO_4}=\dfrac{\dfrac{61}{112}.98}{30\%}=177,92\left(g\right)\\ m_{ddsaupu}=17,6+177,92-\dfrac{61}{11,2}.2=194,43\left(g\right)\\Tacó:\left\{{}\begin{matrix}n_{AlCl_3}=\dfrac{43}{190}\\n_{FeCl_2}=\dfrac{2183}{10640}\end{matrix}\right. \\ C\%_{AlCl_3}=15,54\%;C\%_{FeCl_2}=13,4\%\)
\(a,\) Đặt \(\begin{cases} n_{Al}=x(mol)\\ n_{Fe}=y(mol \end{cases} \Rightarrow 27x+56y=17,6(1)\)
\(n_{H_2}=\dfrac{12,2}{22,4}=0,54(mol)\\ PTHH:2Al+3H_2SO_4\to Al_2(SO_4)_3+3H_2\\ Fe+H_2SO_4\to FeSO_4+H_2\\ \Rightarrow 1,5x+y=0,54(2)\\ (1)(2)\Rightarrow \begin{cases} x=0,22(mol)\\ y=0,21(mol) \end{cases} \Rightarrow \begin{cases} \%_{Al}=\dfrac{0,22.27}{17,6}.100\%=33,75\%\\ \%_{Fe}=100\%-33,75\%=66,25\% \end{cases}\\ b,\Sigma n_{H_2SO_4}=1,5x+y=0,54(mol)\\ \Rightarrow m_{dd_{H_2SO_4}}=\dfrac{0,54.98}{30\%}=176,4(g)\)
\(m_{H_2}=0,54.2=1,08(g)\\ \Rightarrow m_{dd{\text{ sau phản ứng}}}=17,6+176,4-1,08=192,92(g)\\ n_{Al_2(SO_4)_3}=\dfrac{1}{2}n_{Al}=0,11(mol);n_{FeSO_4}=n_{Fe}=0,21(mol)\\ \Sigma m_{\text{các chất sau phản ứng}}=m_{Al_2(SO_4)_3}+m_{FeSO_4}=0,11.342+0,21.152=69,54(g)\\ \Rightarrow C\%_{\text{chất sau phản ứng}}=\dfrac{69,54}{192,92}.100\%=36,05\%\)