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a) \(n_{\left(CH_3COO\right)_2Mg}=\dfrac{1,42}{142}=0,01\left(mol\right)\)
PTHH: Mg + 2CH3COOH --> (CH3COO)2Mg + H2
0,02<-----------0,01-------->0,01
=> VH2 = 0,01.22,4 = 0,224 (l)
\(C_{M\left(CH_3COOH\right)}=\dfrac{0,02}{0,2}=0,1M\)
b)
PTHH: CH3COOH + NaOH --> CH3COONa + H2O
0,02------>0,02
=> \(V_{dd.NaOH}=\dfrac{0,02}{0,2}=0,1\left(l\right)=100\left(ml\right)\)
Ta có: \(n_{Mg}=\dfrac{4,8}{24}=0,2\left(mol\right)\)
PT: \(Mg+2HCl\rightarrow MgCl_2+H_2\)
a, \(n_{H_2}=n_{Mg}=0,2\left(mol\right)\)
\(\Rightarrow V_{H_2}=0,2.22,4=4,48\left(l\right)\)
b, \(n_{HCl}=2n_{H_2}=0,4\left(mol\right)\)
\(\Rightarrow C_{M_{HCl}}=\dfrac{0,4}{0,2}=2\left(M\right)\)
c, PT: \(HCl+KOH\rightarrow KCl+H_2O\)
Theo PT: \(n_{KOH}=n_{HCl}=0,4\left(mol\right)\)
\(\Rightarrow m_{ddKOH}=\dfrac{0,4.56}{5,6\%}=400\left(g\right)\)
\(\Rightarrow V_{ddKOH}=\dfrac{400}{1,045}\approx382,78\left(ml\right)\)
\(n_{\left(CH_3COO\right)_2Mg}=\dfrac{0,71}{142}=0,005\left(mol\right)\)
PTHH: 2CH3COOH + Mg ---> (CH3COO)2Mg + H2
0,01<----------------------0,005---------->0,005
\(C_{M\left(CH_3COOH\right)}=\dfrac{0,01}{0,025}=4M\)
PTHH: CH3COOH + NaOH ---> CH3COONa + H2O
0,005--------->0,005
\(V_{ddNaOH}=\dfrac{0,005}{0,75}=\dfrac{1}{150}M\\ V_{H_2}=0,005.22,4=0,0112\left(l\right)\)
200ml = 0,2l
\(n_{Zn}=\dfrac{19,5}{65}=0,3\left(mol\right)\)
Pt : \(Zn+2HCl\rightarrow ZnCl_2+H_2|\)
1 2 1 1
0,3 0,6 0,3 0,3
a) \(n_{ZnCl2}=\dfrac{0,3.1}{1}=0,3\left(mol\right)\)
\(C_{M_{ZnCl2}}=\dfrac{0,3}{0,2}=1,5\left(M\right)\)
b) \(n_{H2}=\dfrac{0,3.1}{1}=0,3\left(mol\right)\)
\(V_{H2\left(dktc\right)}=0,3.22,4=6,72\left(l\right)\)
c) Pt : \(NaOH+HCl\rightarrow NaCl+H_2O|\)
1 1 1 1
0,6 0,6
\(n_{NaOH}=\dfrac{0,6.1}{1}=0,6\left(mol\right)\)
\(m_{NaOH}=0,6.40=24\left(g\right)\)
\(m_{ddNaOH}=\dfrac{24.100}{20}=120\left(g\right)\)
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\(PTHH:2CH_3COOH+Ca\rightarrow\left(CH_3COO\right)_2Na+H_2\)
Ta có:
\(n_{\left(CH3COO\right)2Na}=\frac{4,47}{158}=0,03\left(mol\right)\)
\(\Rightarrow n_{CH3COOH}=0,03.2=0,06\left(mol\right)\)
\(\Rightarrow CM_{CH3COOH}=\frac{0,06}{0,2}=0,3M\)
\(n_{H2}=0,03\left(mol\right)\Rightarrow V_{H2}=0,03.22,4=6,72\left(l\right)\)
PTHH :
\(CH_3COOH+NaOH\rightarrow CH_3COONa+H_2O\)
_0,06________0,06_____________________
\(\Rightarrow V_{NaOH}=\frac{0,06}{0,5}=0,12\left(l\right)=120\left(ml\right)\)