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1/ nNaCl=5,85/58,5=0,1 mol.
nAgNO3=34/170=0,2 mol.
PTPU: NaCl+AgNO3=>AgCl+NaNO3
vì NaCl và AgNO3 phan ung theo ti le 1:1 (nAgNO3 p.u=nNaCl=0,1 mol)
=>AgNO3 du
nAgNO3 du= 0,2-0,1=0,1 mol.
Ta tinh luong san pham theo chat p.u het la NaCl
sau p.u co: AgNO3 du:0,1 mol; AgCl ket tua va NaCl: nAgCl=nNaNO3=nNaCl=0,1 mol.V(dd)=300+200=500ml=0,5 ()l
=>khoi lg ket tua: mAgCl=0,1.143,5=14,35 g
C(M)AgNO3=C(M)NaNO3=n/V=0,1/0,5=0,2 M
PTHH :
\(2CH3COOH+Mg->\left(CH3COO\right)2Mg+H2\)
0,02mol................0,01mol......................................0,01mol
a) VH2(đktc) = 0,01.22,4 = 0,224(l)
mMg = 0,01.24 = 0,24(g)
b) \(CH3COOH+C2H5OH\underrightarrow{H2SO4,đặc,t0}CH3COOC2H5+H2O\)
Ta có : nC2H5OH = 1,15/46 = 0,025(mol)
Ta có : nC2H5OH = 0,025 > nCH3COOH
=> nC2H5OH còn dư
=> n(etyl axetat) = nCH3COOH = 0,02(mol)
=> m(etyl axetat) = 0,02.88 = 1,76(g)
\(n_{Zn}=\dfrac{4,55}{65}=0,07(mol)\\ Zn+2HCl\to ZnCl_2+H_2\\ a,n_{HCl}=0,14(mol)\\ \Rightarrow C_{M_{HCl}}=\dfrac{0,14}{0,2}=0,7M\\ b,n_{H_2}=0,07(mol)\\ \Rightarrow V_{H_2}=0,07.22,4=1,568(l)\\ c,n_{ZnCl_2}=0,07(mol)\\ \Rightarrow m_{ZnCl_2}=0,07.136=9,52(g)\\ c,ZnCl_2+2AgNO_3\to 2AgCl\downarrow+Zn(NO_3)_2\)
\(m_{dd_{ZnCl_2}}=200.0,8+4,55-0,07.2=164,41(g)\\ n_{AgCl}=0,14(mol);n_{Zn(NO_3)_2}=0,07(mol)\\ \Rightarrow C\%_{Zn(NO_3)_2}=\dfrac{0,07.189}{164,41+200-0,14.143,5}.100\%=3,84%\)
a.
\(C_2H_2+H_2\xrightarrow[Pd]{t^o}C_2H_4\)
\(C_2H_4+H_2\xrightarrow[Ni]{t^o}C_2H_6\)
b.
\(C_2H_4+H_2O\underrightarrow{axit}C_2H_5OH\)
\(C_2H_5OH+O_2\underrightarrow{mengiấm}CH_3COOH+H_2O\)
\(CH_3COOH+C_2H_5OH\xrightarrow[H_2SO_4đặc]{t^o}CH_3COOC_2H_5+H_2O\)
a, \(n_{C_2H_4}=\dfrac{15,68}{22,4}=0,7\left(mol\right)\)
PT: \(C_2H_4+H_2O\underrightarrow{^{t^o,xt}}C_2H_5OH\)
Theo PT: \(n_{C_2H_5OH\left(LT\right)}=n_{C_2H_4}=0,7\left(mol\right)\)
Mà: H = 90%
\(\Rightarrow n_{C_2H_5OH\left(TT\right)}=0,7.90\%=0,63\left(mol\right)\)
\(\Rightarrow m_{C_2H_5OH\left(TT\right)}=0,63.46=28,98\left(g\right)\)
\(\Rightarrow V_{C_2H_5OH}=\dfrac{28,98}{0,8}=36,225\left(ml\right)\)
b, \(C_2H_5OH+O_2\underrightarrow{^{mengiam}}CH_3COOH+H_2O\)
Theo PT: \(n_{CH_3COOH}=n_{C_2H_5OH}=0,63\left(mol\right)\)
\(\Rightarrow m_{CH_3COOH}=0,63.60=37,8\left(g\right)\)
\(\Rightarrow m_{ddCH_3COOH}=\dfrac{37,8}{5\%}=756\left(g\right)\)
Mg + 2HCl -> MgCl2 + H2
0.2 0.4 0.2 0.2
\(nHCl=0.2\times2=0.4mol\)
a.\(m=0.2\times24=4.8g\); \(V=0.2\times22.4=4.48l\)
b.MgCl2 + 2NaOH -> Mg(OH)2 + NaCl
0.2 0.2
\(mNaOH=20\%\times100=20g\Rightarrow nNaOH=0.5mol\)
=> MgCl2 hết, NaOH dư
\(mMg\left(OH\right)2=0.2\times58=11.6g\)