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PTHH\(Na_2CO_3+2HCl\rightarrow2NaCl+H_2O+CO_2\)
tl............1................2.............2.............1.............1..(mol
br 0,1.................0,2......................................0,1(mol)
NaCl không phản ứng đc vsHCl
b)\(n_{CO_2}=\dfrac{2,24}{22,4}=0,1\left(mol\right)\)
\(C_{MHCl}=\dfrac{0,2}{0,4}=0,5\left(M\right)\)(đổi 400ml=0,4(l))
c)\(Tacom_{Na_2CO_3}=0,1.106=10,6\left(g\right)\)
\(\Rightarrow\%m_{Na_2CO_3}=\dfrac{10,6}{20}.100=53\%\)
\(\Rightarrow\%mNaCl=100\%-53\%=47\%\)
Đặt: \(n_{Zn}=a\left(mol\right);n_{ZnO}=b\left(mol\right)\left(a,b>0\right)\)
\(n_{H_2}=\dfrac{2,24}{22,4}=0,1\left(mol\right)\)
\(a.Zn+2HCl\rightarrow ZnCl_2+H_2\\ ZnO+2HCl\rightarrow ZnCl_2+H_2O\\ \Rightarrow\left\{{}\begin{matrix}65a+81b=14,6\\a=0,1\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}a=0,1\\b=0,1\end{matrix}\right.\\ b.m_{Zn}=0,1.65=6,5\left(g\right)\\ m_{ZnO}=0,1.81=8,1\left(g\right)\\ d.m_{ddHCl}=\dfrac{\left(0,1+0,1\right).2.36,5.100}{7,3}=200\left(g\right)\)
a) PTHH : \(Zn+H_2SO_4-->ZnSO_4+H_2\uparrow\) (1)
\(ZnO+H_2SO_4-->ZnSO_4+H_2O\) (2)
b) Theo pthh (1) : \(n_{Zn}=n_{H2}=\dfrac{2,24}{22,4}=0,1\left(mol\right)\)
=> \(m_{Zn}=0,1.65=6,5\left(g\right)\)
=> \(m_{ZnO}=22,7-6,5=16,2\left(g\right)\)
c) \(ZnO=\dfrac{16,2}{81}=0,2\left(mol\right)\)
Theo pthh (1) và (2) : \(\Sigma n_{H2SO4}=n_{Zn}+n_{ZnO}=0,1+0,2=0,3\left(mol\right)\)
=> \(C_{M\left(ddH2SO4\right)}=\dfrac{0,3}{0,1}=1,5M\)
a) PTHH : Zn+H2SO4−−>ZnSO4+H2↑Zn+H2SO4−−>ZnSO4+H2↑ (1)
ZnO+H2SO4−−>ZnSO4+H2OZnO+H2SO4−−>ZnSO4+H2O (2)
b) Theo pthh (1) : nZn=nH2=2,2422,4=0,1(mol)nZn=nH2=2,2422,4=0,1(mol)
=> mZn=0,1.65=6,5(g)mZn=0,1.65=6,5(g)
=> mZnO=22,7−6,5=16,2(g)mZnO=22,7−6,5=16,2(g)
c) ZnO=16,281=0,2(mol)ZnO=16,281=0,2(mol)
Theo pthh (1) và (2) : ΣnH2SO4=nZn+nZnO=0,1+0,2=0,3(mol)ΣnH2SO4=nZn+nZnO=0,1+0,2=0,3(mol)
=> CM(ddH2SO4)=0,30,1=1,5M
tích đúng đê
\(Zn+2HCl->ZnCl_2+H_2\\ Mg+2HCl->MgCl_2+H_2\\ n_{Zn}=a\\ n_{Mg}=b\\ 65a+24b=11,3g\\ n_{H_2}=a+b=\dfrac{6,72}{22,4}=0,3\\ a=0,1\\ m_{Zn}=65.0,1=6,5g\)
a)\(n_{H_2}=\dfrac{3,36}{22,4}=0,15mol\)
\(Zn+2HCl\rightarrow ZnCl_2+H_2\)
0,15 0,3 0,15
\(m_{Zn}=0,15\cdot65=9,75\left(g\right)\)
\(\%m_{Zn}=\dfrac{9,75}{17,85}\cdot100\%=54,62\%\)
\(\%m_{ZnO}=100\%-54,62\%=45,38\%\)
b)\(m_{ZnO}=17,85-9,75=8,1\left(g\right)\Rightarrow n_{ZnO}=\dfrac{8,1}{81}=0,1mol\)
\(ZnO+2HCl\rightarrow ZnCl_2+H_2O\)
0,1 0,2
\(\Rightarrow\Sigma n_{HCl}=0,3+0,2=0,5mol\)
\(\Rightarrow V=\dfrac{0,5}{1}=0,5l=500ml\)
a, \(Zn+2CH_3COOH\rightarrow\left(CH_3COO\right)_2Zn+H_2\)
\(ZnO+2CH_3COOH\rightarrow\left(CH_3COO\right)_2Zn+H_2O\)
b, Ta có: \(n_{H_2}=\dfrac{2,24}{22,4}=0,1\left(mol\right)\)
Theo PT: \(n_{Zn}=n_{H_2}=0,1\left(mol\right)\)
\(\Rightarrow\left\{{}\begin{matrix}\%m_{Zn}=\dfrac{0,1.65}{14,6}.100\%\approx44,52\%\\\%m_{ZnO}\approx55,48\%\end{matrix}\right.\)
c, Ta có: \(n_{ZnO}=\dfrac{14,6-0,1.65}{81}=0,1\left(mol\right)\)
Theo PT: \(n_{CH_3COOH}=2n_{Zn}+2n_{ZnO}=0,4\left(mol\right)\)
PT: \(C_2H_5OH+O_2\underrightarrow{^{mengiam}}CH_3COOH+H_2O\)
Theo PT: \(n_{C_2H_5OH\left(LT\right)}=n_{CH_3COOH}=0,4\left(mol\right)\)
Mà: H = 80%
\(\Rightarrow n_{C_2H_5OH\left(TT\right)}=\dfrac{0,4}{80\%}=0,5\left(mol\right)\)
\(\Rightarrow m_{C_2H_5OH\left(TT\right)}=0,5.46=23\left(g\right)\)
\(\Rightarrow V_{C_2H_5OH}=\dfrac{23}{0,8}=28,75\left(ml\right)\)
\(\Rightarrow V_{C_2H_5OH\left(18,4^o\right)}=\dfrac{28,75}{18,4}.100=156,25\left(ml\right)=0,15625\left(l\right)\)
Câu 1
\(a)PTHH:Mg+2HCl\rightarrow MgCl_2+H_2\\
b)200ml=0,2l\\
n_{HCl}=0,2.1=0,2mol\\
n_{H_2}=n_{MgCl_2}=\dfrac{1}{2}n_{H_2}=\dfrac{1}{2}\cdot0,2=0,1mol\\
V_{H_2}=0,1.24,79=2,479l\\
c)C_{M_{MgCl_2}}=\dfrac{0,1}{0,2}=0,5M\)
a) PTHH: \(Zn+H_2SO_4\rightarrow ZnSO_4+H_2\uparrow\) (1)
\(ZnO+H_2SO_4\rightarrow ZnSO_4+H_2O\) (2)
b) Ta có: \(\left\{{}\begin{matrix}n_{H_2}=\dfrac{2,24}{22,4}=0,1\left(mol\right)\\\Sigma n_{H_2SO_4}=0,2\cdot1=0,2\left(mol\right)\end{matrix}\right.\)
\(\Rightarrow n_{Zn}=n_{ZnO}=n_{H_2SO_4\left(1\right)}=n_{H_2SO_4\left(2\right)}=0,1\left(mol\right)\)
\(\Rightarrow\left\{{}\begin{matrix}m_{Zn}=0,1\cdot65=6,5\left(g\right)\\m_{ZnO}=0,1\cdot81=8,1\left(g\right)\end{matrix}\right.\)
c) Theo các PTHH: \(\left\{{}\begin{matrix}n_{Zn}=n_{ZnSO_4\left(1\right)}=0,1mol\\n_{ZnO}=n_{ZnSO_4\left(2\right)}=0,1mol\end{matrix}\right.\)
\(\Rightarrow\Sigma n_{ZnSO_4}=0,2mol\) \(\Rightarrow C_{M_{ZnSO_4}}=\dfrac{0,2}{0,2}=1\left(M\right)\)
(Coi như thể tích dd thay đổi không đáng kể)