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Gọi \(\left\{{}\begin{matrix}n_{Mg}=x\\n_{Zn}=y\end{matrix}\right.\)
\(n_{H_2}=\dfrac{6,72}{22,4}=0,3mol\)
\(Mg+2CH_3COOH\rightarrow\left(CH_3COO\right)_2Mg+H_2\)
x 2x x x ( mol )
\(Zn+2CH_3COOH\rightarrow\left(CH_3COO\right)_2Zn+H_2\)
y 2y y y ( mol )
Ta có:
\(\left\{{}\begin{matrix}24x+65y=11,3\\x+y=0,3\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}x=0,2\\y=0,1\end{matrix}\right.\)
\(\rightarrow\left\{{}\begin{matrix}m_{Mg}=0,2.24=4,8g\\m_{Zn}=0,1.65=6,5g\end{matrix}\right.\)
\(\rightarrow\left\{{}\begin{matrix}\%m_{Mg}=\dfrac{4,8}{11,3}.100=42,47\%\\\%m_{Zn}=100\%-42,47\%=57,53\%\end{matrix}\right.\)
\(m_{CH_3COOH}=60.\left(0,2+0,1\right)=18g\)
\(C\%_{CH_3COOH}=\dfrac{18}{200}.100=9\%\)
\(\left\{{}\begin{matrix}m_{\left(CH_3COO\right)_2Mg}=0,2.142=28,4g\\m_{\left(CH_3COO\right)_2Zn}=0,1.183=18,3g\end{matrix}\right.\)
\(m_{ddspứ}=11,3+200-0,3.2=210,7g\)
\(\rightarrow\left\{{}\begin{matrix}C\%_{\left(CH_3COO\right)_2Mg}=\dfrac{28,4}{210,7}.100=13,47\%\\C\%_{\left(CH_3COO\right)_2Zn}=\dfrac{18,3}{210,7}.100=8,68\%\end{matrix}\right.\)
Gọi \(\left\{{}\begin{matrix}n_{Mg}=a\left(mol\right)\\n_{Zn}=b\left(mol\right)\end{matrix}\right.\)
\(n_{H_2}=\dfrac{4,48}{22,4}=0,2\left(mol\right)\)
PTHH:
Mg + 2CH3COOH ---> (CH3COO)2Mg + H2
a---->2a-------------------->a------------------>a
Zn + 2CH3COOH ---> (CH3COO)2Zn + H2
b---->2b------------------->b------------------>b
=> \(\left\{{}\begin{matrix}24a+65b=8,9\\a+b=0,2\end{matrix}\right.\Leftrightarrow a=b=0,1\left(mol\right)\)
=> \(\left\{{}\begin{matrix}\%m_{Mg}=\dfrac{0,1.24}{8,9}.100\%=27\%\\\%m_{Zn}=100\%-27\%=73\%\end{matrix}\right.\)
=> \(C\%_{CH_3COOH}=\dfrac{\left(0,1.2+0,1.2\right).60}{200}.100\%=12\%\)
\(m_{dd}=200+8,9-0,2.2=208,5\left(g\right)\)
=> \(\left\{{}\begin{matrix}C\%_{\left(CH_3COO\right)_2Mg}=\dfrac{0,1.142}{208,5}.100\%=6,81\%\\C\%_{\left(CH_3COO\right)_2Zn}=\dfrac{0,1.183}{208,5}.100\%=8,78\%\end{matrix}\right.\)
\(n_{H_2}=\dfrac{4,48}{22,4}=0,2mol\)
Gọi \(\left\{{}\begin{matrix}n_{Mg}=x\\n_{Zn}=y\end{matrix}\right.\) ( mol )
\(\rightarrow24x+65y=8,9\left(g\right)\) (1)
\(Mg+2CH_3COOH\rightarrow\left(CH_3COO\right)_2Mg+H_2\uparrow\)
x 2x x x ( mol )
\(Zn+2CH_3COOH\rightarrow\left(CH_3COO\right)_2Zn+H_2\uparrow\)
y 2y y y ( mol )
\(\rightarrow x+y=0,2\left(mol\right)\) (2)
\(\left(1\right);\left(2\right)\rightarrow\left\{{}\begin{matrix}x=0,1\\y=0,1\end{matrix}\right.\)
\(\rightarrow\left\{{}\begin{matrix}\%m_{Mg}=\dfrac{0,1.24}{8,9}.100\%=26,96\%\\\%m_{Zn}=100\%-26,96\%=73,04\%\end{matrix}\right.\)
\(m_{CH_3COOH}=\left(2.0,1+2.0,1\right).60=24g\)
\(C\%_{CH_3COOH}=\dfrac{24}{200}.100\%=12\%\)
\(\left\{{}\begin{matrix}m_{\left(CH_3COO\right)_2Mg}=0,1.142=14,2g\\m_{\left(CH_3COO\right)_2Zn}=0,1.183=18,3g\end{matrix}\right.\)
a, nZnO= 16,2/81= 0,2(mol)
PTHH
ZnO + 2CH3COOH -> (CH3COO)2Zn + H2O
b,theo pt ta có:
n(CH3COO)2Zn=nZnO=0,2mol
-> m(CH3COO)2Zn= 0,2 . 183= 36,6(g)
c,mình chưa làm được nha :)
a) Fe + 2HCl --> FeCl2 + H2
b) \(n_{Fe}=\dfrac{22,4}{56}=0,4\left(mol\right)\)
PTHH: Fe + 2HCl --> FeCl2 + H2
_____0,4--->0,8------>0,4--->0,4
=> VH2 = 0,4.22,4 = 8,96(l)
c) mHCl = 0,8.36,5 = 29,2 (g)
=> \(m_{dd\left(HCl\right)}=\dfrac{29,2.100}{7,3}=400\left(g\right)\)
mdd (sau pư) = 22,4 + 400 - 0,4.2 = 421,6 (g)
=> \(C\%\left(FeCl_2\right)=\dfrac{127.0,4}{421,6}.100\%=12,05\%\)
\(n_{Al}=\dfrac{10,8}{27}=0,4(mol)\\ a,2Al+3H_2SO_4\to Al_2(SO_4)_3+3H_2\\ \Rightarrow n_{H_2}=n_{H_2SO_4}=0,6(mol);n_{Al_2(SO_4)_3}=0,2(mol)\\ b,V_{H_2}=0,6.22,4=13,44(l)\\ c,m_{dd_{H_2SO_4}}=\dfrac{0,6.98}{9,8\%}=600(g)\\ \Rightarrow C\%_{Al_2(SO_4)_3}=\dfrac{0,2.342}{10,8+600-0,6.2}.100\%=11,22\%\)
\(Fe+2CH_3COOH\rightarrow\left(CH_3COO\right)_2Fe+H_2\)
\(n_{Fe}=\dfrac{11,2}{56}=0,2\left(mol\right)\)
Theo PT: \(n_{\left(CH_3COO\right)_2Fe}=n_{Fe}=0,2\left(mol\right)\Rightarrow m_{\left(CH_3COO\right)_2Fe}=0,2.174=34,8\left(g\right)\)
Ta có: \(n_{H_2}=n_{Fe}=0,2\left(mol\right)\)
m dd sau pư = 11,2 + 200 - 0,2.2 = 210,8 (g)
\(\Rightarrow C\%_{\left(CH_3COO\right)_2Fe}=\dfrac{34,8}{210,8}.100\%\approx16,51\%\)
Bài 4 :
\(n_{Zn}=\dfrac{13}{65}=0,2\left(mol\right)\)
a) Pt : \(Zn+2HCl\rightarrow ZnCl_2+H_2|\)
1 2 1 1
0,2 0,4 0,2 0,2
b) \(n_{H2}=\dfrac{0,2.1}{1}=0,2\left(mol\right)\)
\(V_{H2\left(dktc\right)}=0,2.22,4=4,48\left(l\right)\)
c) \(n_{HCl}=\dfrac{0,2.2}{1}=0,4\left(mol\right)\)
⇒ \(m_{HCl}=0,4.36,5=14,6\left(g\right)\)
\(C_{ddHCl}=\dfrac{14,6.100}{100}=14,6\)0/0
d) \(n_{ZnCl2}=\dfrac{0,2.1}{1}=0,2\left(mol\right)\)
⇒ \(m_{ZnCl2}=0,2.136=27,2\left(g\right)\)
\(m_{ddspu}=13+100-\left(0,2.2\right)=112,6\left(g\right)\)
\(C_{ZnCl2}=\dfrac{27,2.100}{112,6}=24,16\)0/0
Chúc bạn học tốt
Bài 3 :
\(n_{Mg}=\dfrac{12}{24}=0,5\left(mol\right)\)
a) Pt : \(Mg+H_2SO_4\rightarrow MgSO_4+H_2|\)
1 1 1 1
0,5 0,5 0,5 0,5
b) \(n_{H2}=\dfrac{0,5.1}{1}=0,5\left(mol\right)\)
\(V_{H2\left(dktc\right)}=0,5.22,4=11,2\left(l\right)\)
c) \(n_{H2SO4}=\dfrac{0,5.1}{1}=0,5\left(mol\right)\)
⇒ \(m_{H2SO4}=0,5.98=49\left(g\right)\)
\(C_{ddH2SO4}=\dfrac{49.100}{200}=24,5\)0/0
d) \(n_{MgSO4}=\dfrac{0,5.1}{1}=0,5\left(mol\right)\)
⇒ \(m_{MgSO4}=0,5.120=60\left(g\right)\)
\(m_{ddspu}=12+200-\left(0,5.2\right)=211\left(g\right)\)
\(C_{MgSO4}=\dfrac{60.100}{211}=28,44\)0/0
Chúc bạn học tốt
\(n_{MgCO3}=\dfrac{8,4}{84}=0,1\left(mol\right)\)
a) Pt : \(2CH_3COOH+MgCO_3\rightarrow\left(CH_3COO\right)_2Mg+CO_2+H_2O|\)
2 1 1 1 1
0,2 0,1 0,1 0,1
b) \(n_{CH3COOH}=\dfrac{0,1.2}{1}=0,2\left(mol\right)\)
⇒ \(m_{CH3COOH}=0,2.60=12\left(g\right)\)
\(C_{ddCH3COOH}=\dfrac{12.100}{200}=6\)0/0
\(n_{\left(CH3COO\right)2Mg}=\dfrac{0,1.1}{1}=0,1\left(mol\right)\)
⇒ \(m_{\left(CH3COO\right)2Mg}=0,1.142=14,2\left(g\right)\)
\(m_{ddspu}=8,4+200-\left(0,1.44\right)=204\left(g\right)\)
\(C_{dd\left(CH3COO\right)2Mg}=\dfrac{14,2.100}{204}=6,96\)0/0
Chúc bạn học tốt