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a,\(n_{Fe}=\dfrac{5,6}{56}=0,1\left(mol\right)\)
PTHH: Fe + H2SO4 → FeSO4 + H2
Mol: 0,1 0,1 0,1 0,1
b,\(V_{H_2}=0,1.22,4=2,24\left(l\right)\)
c,\(C_{M_{ddH_2SO_4}}=\dfrac{0,1}{0,2}=0,5M\)
d,\(C_{M_{ddFeSO_4}}=\dfrac{0,1}{0,2}=0,5M\)
\(n_{H2}=\dfrac{2,24}{22,4}=0,1\left(mol\right)\)
a) \(Fe+H_2SO_4\rightarrow FeSO_4+H_2\)
\(Al_2O_3+3H_2SO_4\rightarrow Al_2\left(SO_4\right)_3+3H_2O\)
b) \(n_{Fe}=n_{H2}=n_{H2SO4}=0,1\left(mol\right)\Rightarrow m_{Fe}=0,1.56=5,6\left(g\right)\)
\(\Rightarrow m_{Al2O3}=15,8-5,6=10,2\left(g\right)\)
c) Ta có : \(n_{Al2O3}=\dfrac{10,2}{102}=0,1\left(mol\right)\Rightarrow n_{H2SO4}=3n_{Al2O3}=0,3\left(mol\right)\)
\(C_{MddH2SO4}=\dfrac{0,1+0,3}{0,2}=2M\)
\(n_{Fe}=\dfrac{5,6}{56}=0,1\left(mol\right)\)
PTHH: Fe + H2SO4 → FeSO4 + H2
Mol: 0,1 0,1 0,1 0,1
\(V_{H_2}=0,1.24,79=2,479\left(l\right)\)
\(C_{M_{ddH_2SO_4}}=\dfrac{0,1}{0,2}=0,5M\)
\(C_{M_{ddFeSO_4}}=\dfrac{0,1}{0,2}=0,5M\)
\(n_{Fe}=\dfrac{5,6}{56}=0,1\left(mol\right)\)
Pt : \(Fe+H_2SO_4\rightarrow FeSO_4+H_2|\)
1 1 1 1
0,1 0,1 0,1
a) \(n_{H2SO4}=\dfrac{0,1.1}{1}=0,1\left(mol\right)\)
\(V_{ddH2SO4}=\dfrac{0,1}{2}=0,05\left(l\right)\)
b) \(n_{FeSO4}=\dfrac{0,1.1}{1}=0,1\left(mol\right)\)
\(C_{M_{FeSO4}}=\dfrac{0,1}{0,05}=2\left(M\right)\)
Chúc bạn học tốt
a) PTHH: \(Zn+H_2SO_4\rightarrow ZnSO_4+H_2\uparrow\)
b+c)
Ta có: \(n_{Zn}=\dfrac{19,5}{65}=0,3\left(mol\right)=n_{H_2SO_4}=n_{H_2}\)
\(\Rightarrow\left\{{}\begin{matrix}V_{H_2}=0,3\cdot22,4=6,72\left(l\right)\\m_{ddH_2SO_4}=\dfrac{0,3\cdot98}{30\%}=98\left(g\right)\end{matrix}\right.\)
d) PTHH: \(ZnSO_4+BaCl_2\rightarrow ZnCl_2+BaSO_4\downarrow\)
Ta có: \(n_{BaCl_2}=\dfrac{260\cdot20\%}{208}=0,25\left(mol\right)\)
Xét tỉ lệ: \(\dfrac{0,3}{1}>\dfrac{0,25}{1}\) \(\Rightarrow\) ZnSO4 còn dư, BaCl2 phản ứng hết
\(\Rightarrow\left\{{}\begin{matrix}n_{ZnCl_2}=0,25mol=n_{BaSO_4}\\n_{ZnSO_4\left(dư\right)}=0,05mol\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}m_{ZnCl_2}=0,25\cdot136=34\left(g\right)\\m_{BaSO_4}=0,25\cdot233=58,25\left(g\right)\\m_{ZnSO_4\left(dư\right)}=0,05\cdot161=8,05\left(g\right)\end{matrix}\right.\)
Mặt khác: \(m_{H_2}=0,3\cdot2=0,6\left(g\right)\)
\(\Rightarrow m_{dd}=m_{Zn}+m_{ddH_2SO_4}-m_{H_2}+m_{ddBaCl_2}-m_{BaSO_4}=318,65\left(g\right)\)
\(\Rightarrow\left\{{}\begin{matrix}C\%_{ZnCl_2}=\dfrac{34}{318,65}\cdot100\%\approx10,67\%\\C\%_{ZnSO_4\left(dư\right)}=\dfrac{8,05}{318,65}\cdot100\%\approx2,53\%\end{matrix}\right.\)
\(n_{H_2}=\dfrac{3,24}{24}=0,135(mol)\\ Fe+H_2SO_4\to FeSO_4+H_2\\ \Rightarrow n_{H_2SO_4}=n_{FeSO_4}=0,135(mol)\\ \Rightarrow \begin{cases} C_{M_{H_2SO_4}}=\dfrac{0,135}{0,2}=0,675M\\ C_{M_{FeSO_4}}=\dfrac{0,135}{0,2}=0,675M \end{cases}\)
Bài 14 :
\(n_{Fe}=\dfrac{8,4}{56}=0,15\left(mol\right)\)
Pt : \(Fe+2HCl\rightarrow FeCl_2+H_2|\)
1 2 1 1
0,15 0,3 0,15 0,15
a) \(n_{H2}=\dfrac{0,15.1}{1}=0,15\left(mol\right)\)
\(V_{H2\left(dktc\right)}=0,15.22,4=3,36\left(l\right)\)
b) \(n_{HCl}=\dfrac{0,15.2}{1}=0,3\left(mol\right)\)
\(V_{ddHCl}=\dfrac{0,3}{2}=0,15\left(l\right)\)
c) \(n_{FeCl2}=\dfrac{0,15.1}{1}=0,15\left(mol\right)\)
\(C_{M_{FeCl2}}=\dfrac{0,15}{0,15}=1\left(M\right)\)
Chúc bạn học tốt
ở đoạn c bạn có ghi nhầm ko à , tại mình cứ thấy nó sai sai
\(n_{Al}=\dfrac{10,8}{27}=0,4(mol)\\ a,2Al+3H_2SO_4\to Al_2(SO_4)_3+3H_2\\ \Rightarrow n_{H_2}=n_{H_2SO_4}=0,6(mol);n_{Al_2(SO_4)_3}=0,2(mol)\\ b,V_{H_2}=0,6.22,4=13,44(l)\\ c,m_{dd_{H_2SO_4}}=\dfrac{0,6.98}{9,8\%}=600(g)\\ \Rightarrow C\%_{Al_2(SO_4)_3}=\dfrac{0,2.342}{10,8+600-0,6.2}.100\%=11,22\%\)
a) Fe + H2SO4 --> FeSO4 + H2
b) \(n_{Fe}=\dfrac{5,6}{56}=0,1\left(mol\right)\)
PTHH: Fe + H2SO4 --> FeSO4 + H2
0,1-->0,1--------------->0,1
=> \(C_{M\left(dd.H_2SO_4\right)}=\dfrac{0,1}{0,1}=1M\)
c) VH2 = 0,1.22,4 = 2,24 (l)
kia là 100 mililít hay lít hả bn ?