Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
\(n_{Fe}=\dfrac{23,2}{232}=0,1\left(mol\right)\\ n_{H_2SO_4}=\dfrac{200.29,4}{100}:36,5\approx1,6\left(mol\right)\\ Fe_3O_4+8HCl\xrightarrow[]{}2FeCl_3+FeCl_2+4H_2O\\ \Rightarrow\dfrac{0,1}{1}< \dfrac{1,6}{8}\Rightarrow HCl.dư\\ n_{FeCl_3}=0,1.2=0,2\left(mol\right)\\ n_{FeCl_2}=n_{Fe_3O_4}=0,1mol\\ n_{HCl\left(dư\right)}=1,6-\left(0,1.8\right)=0,8\left(mol\right)\\ m_{dd}=200+23,2=223,2\left(g\right)\\ C_{\%FeCl_3}=\dfrac{0,2.162,5}{223,2}\cdot100\approx14,55\%\\ C_{\%FeCl_2}=\dfrac{0,1.127}{223,2}\cdot100\approx5,67\%\\ C_{HCl\left(dư\right)}=\dfrac{0,8.36,5}{223,2}\cdot100\approx13,08\%\)
\(a)n_{H_2SO_4}=\dfrac{58,8.20}{100.98}=0,12mol\\ n_{BaCl_2}=\dfrac{200.5,2}{100.208}=0,05mol\\ BaCl_2+H_2SO_4\rightarrow BaSO_4+2HCl\\ \Rightarrow\dfrac{0,12}{1}>\dfrac{0,05}{2}\Rightarrow H_2SO_4.dư\\ BaCl_2+H_2SO_4\rightarrow BaSO_4+2HCl\)
0,05 0,05 0,05 0,1
\(m_{BaSO_4}=0,05.233=11,65g\\ b)m_{dd}=58,8+200-11,65=247,15g\\ C_{\%HCl}=\dfrac{0,1.36,5}{247,15}\cdot100=1,48\%\\ C_{\%H_2SO_4,dư}=\dfrac{\left(0,12-0,05\right).98}{247,15}\cdot100=2,78\%\)
a, \(Zn+H_2SO_4\rightarrow ZnSO_4+H_2\)
b, Ta có: \(n_{H_2}=\dfrac{2,24}{22,4}=0,1\left(mol\right)\)
Theo PT: \(n_{Zn}=n_{H_2}=0,1\left(mol\right)\)
\(\Rightarrow\left\{{}\begin{matrix}\%m_{Zn}=\dfrac{0,1.65}{10,5}.100\%\approx61,9\%\\\%m_{Cu}\approx38,1\%\end{matrix}\right.\)
c, \(n_{H_2SO_4}=n_{H_2}=0,1\left(mol\right)\)
\(\Rightarrow C_{M_{H_2SO_4}}=\dfrac{0,1}{0,5}=0,2\left(M\right)\)
a, \(Fe+H_2SO_4\rightarrow FeSO_4+H_2\)
\(n_{H_2}=\dfrac{7,437}{24,79}=0,3\left(mol\right)\)
Theo PT: \(n_{Fe}=n_{H_2}=0,3\left(mol\right)\Rightarrow m_{Fe}=0,3.56=16,8\left(g\right)\)
b, \(n_{H_2SO_4}=n_{H_2}=0,3\left(mol\right)\Rightarrow C\%_{H_2SO_4}=\dfrac{0,3.98}{120}.100\%=24,5\%\)
c, m dd sau pư = 16,8 + 120 - 0,3.2 = 136,2 (g)
d, \(n_{FeSO_4}=n_{H_2}=0,3\left(mol\right)\)
\(\Rightarrow C\%_{FeSO_4}=\dfrac{0,3.152}{136,2}.100\%\approx33,48\%\)
1 , \(n_{Na}=\frac{4,6}{23}=0,2\left(mol\right)\)
\(m_{HCl}=200.2,92\%=5,84\left(mol\right)\) => \(n_{HCl}=\frac{5,84}{36,5}=0,16\left(mol\right)\)
\(2Na+2HCl->2NaCl+H_2\left(1\right)\)
vì \(\frac{0,2}{2}>\frac{0,16}{2}\) => Na dư , HCl hết
dung dịch thu được là dung dịch NaCl
theo (1) \(n_{NaCl}=n_{HCl}=0,16\left(mol\right)\) => \(m_{NaCl}=0,16.58,5=9,36\left(g\right)\)
\(n_{H_2}=\frac{1}{2}n_{HCl}=0,08\left(mol\right)\)
khối lượng dung dịch sau phản ứng là
4,6+200-0,08.2=204,44(g)
\(C_{\%\left(NaCl\right)}=\frac{9,36}{204,44}.100\%\approx4,58\%\)
\(n_{Na}=\dfrac{m}{M}=\dfrac{4,6}{23}=0,2\left(mol\right)\)
\(n_{ddH_2SO_4}=\dfrac{m}{M}=\dfrac{200}{98}=2\)
PTHH:\(2Na+H_2SO_4\rightarrow Na_2SO_4+H_2\)
tpư: 0,2 2
pư: 0,2 0,1 0,1 0,1
spư: 0 1,9 0,1 0,1
a)\(V_{H_2}=n.22,4\)=0,1.22,4=2,24
b)\(m_{Na_2SO_4}=n.M\)=0,1.142=14,2
\(m_{H_2SO_4dư}=n.M\)=1,9.98=186,2
c)\(C\%H_2SO_4=\dfrac{m_{ct}}{m_{dd}}.100=\dfrac{0,1.98}{200}.100\)=0,099%
\(n_{BaCl_2}=\dfrac{52}{208}=0,25(mol)\\ BaCl_2+H_2SO_4\to BaSO_4\downarrow+2HCl\\ \Rightarrow n_{BaSO_4}=0,25(mol);n_{HCl}=0,5(mol)\\ \Rightarrow C\%_{HCl}=\dfrac{0,5.36,5}{52+150-0,25.233}.100\%=12,696\%\)
Ta có: \(n_{BaCl_2}=\dfrac{52}{208}=0,25\left(mol\right)\)
\(PTHH:BaCl_2+H_2SO_4--->BaSO_4\downarrow+2HCl\)
Theo PT: \(n_{HCl}=2.n_{BaCl_2}=2.0,25=0,5\left(mol\right)\)
\(\Rightarrow m_{HCl}=0,5.36,5=18,25\left(g\right)\)
Theo PT: \(n_{BaSO_4}=n_{BaCl_2}=0,25\left(mol\right)\)
\(\Rightarrow m_{BaSO_4}=0,25.233=58,25\left(g\right)\)
\(\Rightarrow m_{dd_{HCl}}=52+150-58,25=143,75\left(g\right)\)
\(\Rightarrow C_{\%_{HCl}}=\dfrac{18,25}{143,75}.100\%=12,7\%\)