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\(n_{H_2}=\dfrac{4,48}{22,4}=0,2\left(mol\right)\)
PTHH: Zn + 2HCl ---> ZnCl2 + H2
0,2<--0,4<------0,2<-----0,2
=> \(\left\{{}\begin{matrix}\%m_{Zn}=\dfrac{0,2.65}{21,1}.100\%=61,61\%\\\%m_{ZnO}=100\%-61,61\%=38,39\%\end{matrix}\right.\)
\(n_{ZnO}=\dfrac{21,1-0,2.65}{81}=0,1\left(mol\right)\)
PTHH: ZnO + 2HCl ---> ZnCl2 + H2O
0,1---->0,2------>0,1
=> \(C\%_{HCl}=\dfrac{\left(0,2+0,4\right).36,5}{200}.100\%=10,95\%\)
\(m_{mu\text{ố}i}=m_{ZnCl_2}=\left(0,1+0,2\right).136=40,8\left(g\right)\)
a) Gọi số mol CH3COOH, C2H5OH là a, b (mol)
=> 60a + 46b = 25,8 (1)
\(n_{H_2}=\dfrac{5,6}{22,4}=0,25\left(mol\right)\)
PTHH: 2Na + 2CH3COOH --> 2CH3COONa + H2
a------------------------->0,5a
2Na + 2C2H5OH --> 2C2H5ONa + H2
b--------------------->0,5b
=> 0,5a + 0,5b = 0,25 (2)
(1)(2) => a = 0,2 (mol); b = 0,3 (mol)
=> \(\left\{{}\begin{matrix}\%m_{CH_3COOH}=\dfrac{0,2.60}{25,8}.100\%=46,51\%\\\%m_{C_2H_5OH}=\dfrac{0,3.46}{25,8}.100\%=53,49\%\end{matrix}\right.\)
b)
\(n_{CH_3COOC_2H_5}=\dfrac{13,2}{88}=0,15\left(mol\right)\)
PTHH: CH3COOH + C2H5OH --H2SO4(đ),to--> CH3COOC2H5 + H2O
Xét tỉ lệ: \(\dfrac{0,2}{1}< \dfrac{0,3}{1}\) => Hiệu suất tính theo CH3COOH
PTHH: CH3COOH + C2H5OH --H2SO4(đ),to--> CH3COOC2H5 + H2O
0,15<---------------------------------0,15
=> \(H=\dfrac{0,15}{0,2}.100\%=75\%\)
\(n_{H_2}=\dfrac{13,44}{22,4}=0,6\left(mol\right)\)
\(2A1+2NAOH+2H_2O-2NaA10_2+H_2O\)
\(AI_2O_3=2NaOH+2NaOHA10_2+H_2O\)
\(n_{AI}=\dfrac{2}{3}n_{H_2}=\dfrac{2}{3}.0,6=0,4\left(mol\right)\)
\(m_{AI}=27.0,4=10,8\left(gam\right);mAI_2O_3=31,2-10,8=20,4\left(gam\right)\)
Biết làm mỗi câu A
a) PTHH: 2Al + 3H2SO4 --> Al2(SO4)3 + 3H2
b) \(n_{H_2}=\dfrac{0,672}{22,4}=0,03\left(mol\right)\)
PTHH: 2Al + 3H2SO4 --> Al2(SO4)3 + 3H2
_____0,02<---0,03<---------------------0,03
=> \(\left\{{}\begin{matrix}\%Al=\dfrac{0,02.27}{2,16}.100\%=25\%\\\%Cu=100\%-25\%=75\%\end{matrix}\right.\)
c) mH2SO4 = 0,03.98 = 2,94 (g)
=> \(C\%\left(H_2SO_4\right)=\dfrac{2,94}{200}.100\%=1,47\%\)
Gọi a, b lần lượt là mol của Al và Zn
\(2Al+6HCl\rightarrow2AlCl_3+3H_2\)
a 1,5a
\(Zn+2HCl\rightarrow ZnCl_2+H_2\)
b b
\(\Rightarrow\left\{{}\begin{matrix}27a+65b=9,2\\1,5a+b=\dfrac{5,6}{22,4}\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}a=0,1\\b=0,1\end{matrix}\right.\)
\(\%m_{Al}=\dfrac{0,1.27}{9,2}.100\%=29,35\%\)
\(\%m_{Zn}=\dfrac{0,1.65}{9,2}.100\%=70,35\%\)
b. \(n_{H_2}=0,25mol\) \(\Rightarrow n_{HCl}=0,5mol\)
\(\Rightarrow m_{HCl}=0,5.36,5=18,25g\)
Ta có: \(10\%=\dfrac{18,25}{m_{dd}}.100\%\)
\(\Leftrightarrow m_{dd}=182,5g\)
a, Gọi \(\left\{{}\begin{matrix}n_{C_2H_5OH}=a\left(mol\right)\\n_{CH_3COOH}=b\left(mol\right)\end{matrix}\right.\)
\(n_{H_2}=\dfrac{11,2}{22,4}=0,5\left(mol\right)\)
PTHH:
2C2H5OH + 2Na ---> 2C2H5ONa + H2
a---------------------------------------->0,5a
2CH3COOH + 2Na ---> 2CH3COONa + H2
b------------------------------------------------>0,5b
=> hệ pt \(\left\{{}\begin{matrix}46a+60b=48,8\\0,5a+0,5b=0,5\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}a=0,8\left(mol\right)\\b=0,2\left(mol\right)\end{matrix}\right.\)
\(\Rightarrow\left\{{}\begin{matrix}m_{C_2H_5OH}=0,8.46=36,8\left(g\right)\\m_{CH_3COOH}=0,2.60=12\left(g\right)\end{matrix}\right.\\ \Rightarrow\left\{{}\begin{matrix}\%m_{C_2H_5OH}=\dfrac{36,8}{48,8}.100\%=75,41\%\\\%m_{CH_3COOH}=100\%-75,41\%=24,59\%\end{matrix}\right.\)
b, PTHH:
\(C_2H_5OH+CH_3COOH\xrightarrow[t^o]{H_2SO_4đặc}CH_3COOC_2H_5+H_2O\)
LTL: 0,8 > 0,2 => Rượu dư
\(n_{CH_3COOC_2H_5\left(tt\right)}=0,2.85\%=0,17\left(mol\right)\\ m_{este}=0,17.88=14,96\left(g\right)\)
a.Gọi \(\left\{{}\begin{matrix}n_{C_2H_5OH}=x\\n_{CH_3COOH}=y\end{matrix}\right.\)
\(n_{H_2}=\dfrac{11,2}{22,4}=0,5mol\)
\(2C_2H_5OH+2Na\rightarrow2C_2H_5ONa+H_2\)
x 1/2 x ( mol )
\(2CH_3COOH+Na\rightarrow2CH_3COONa+H_2\)
y 1/2 y ( mol )
Ta có:
\(\left\{{}\begin{matrix}46x+60y=48,8\\\dfrac{1}{2}x+\dfrac{1}{2}y=0,5\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}x=0,8\\y=0,2\end{matrix}\right.\)
\(\rightarrow m_{C_2H_5OH}=0,8.46=36,8g\)
\(\rightarrow\left\{{}\begin{matrix}\%m_{C_2H_5OH}=\dfrac{36,8}{48,8}.100=75,4\%\\\%m_{CH_3COOH}=100\%-75,4\%=24,6\%\end{matrix}\right.\)
b.\(C_2H_5OH+CH_3COOH\rightarrow\left(H_2SO_4\left(đ\right),t^o\right)CH_3COOC_2H_5+H_2O\)
0,8 < 0,2 ( mol )
0,2 0,2 ( mol )
\(m_{CH_3COOC_2H_5}=0,2.88.85\%=14,96g\)
Đặt :
nC2H5OH = a (mol)
nCH3OH = b (mol)
=> mhh = 46a + 32b = 5.5 (g) (1)
nH2 = 0.075 (mol)
C2H5OH + K => C2H5OK + 1/2 H2
CH3OH + K => CH3OK + 1/2H2
=> 0.5a + 0.5b = 0.075 (2)
(1) , (2) :
a = 0.05
b = 0.1
%m C2H5OH = 0.05*46/5.5 * 100% = 41.82%
%m CH3OH = 58.18%