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nZn=0,1 mol
Zn +2HCl=> ZnCl2+ H2
0,1 mol =>0,2 mol
=>mHCl=36,5.0,2=7,3g
=>m dd HCl=7,3/14,6%=50g
mdd sau pứ=6,5+50-0,1.2=56,3g
=>C% dd ZnCl2=(0,1.136)/56,3.100%=24,16%
a.b. Zn + 2HCl ---> ZnCl2 + H2 (1)
Theo pt: 65g 73g 136g 2g
Theo đề: 6,5g 7,3g 13,6g
=> mddHCl=\(\frac{7,3.100}{14,6}=50\left(g\right)\)
c. Từ pt (1), ta có: \(C_{\%}=\frac{13,6}{50+6,5}.100\%=24,1\%\)
\(n_{Fe}=\dfrac{11,2}{56}=0,2\left(mol\right)\)
\(PTHH:Fe+2HCl\rightarrow FeCl_2+H_2\)
(mol)____0,2____0,4____0,2____0,2__
\(a.m_{FeCl_2}=0,2.127=25,4\left(g\right)\)
\(b.C\%_{ddFeCl_2}=\dfrac{m_{ct}}{m_{ddspu}}.100=\dfrac{25,4}{11,2+120-0,2.2}.100=19,4\left(\%\right)\)
\(c.C\%_{ddHCl}=\dfrac{36,5.0,4}{120}.100=12,17\left(\%\right)\)
\(a,n_{H_2}=\dfrac{1,12}{22,4}=0,05\left(mol\right)\)
PTHH: \(Fe+2HCl\rightarrow FeCl_2+H_2\uparrow\)
0,05<-0,1<------0,05<---0,05
\(b,m_{Fe}=0,05.56=2,8\left(g\right)\\ c,m_{ddHCl}=\dfrac{0,1.36,5}{14,6\%}=25\left(g\right)\\ m_{dd}=25+2,8-0,05.2=27,7\left(g\right)\\ \rightarrow C\%_{FeCl_2}=\dfrac{0,05.127}{27,7}.100\%=22,92\%\)
\(n_{HCl}=\dfrac{400\cdot36.5\%}{36.5}=4\left(mol\right)\)
\(Mg+2HCl\rightarrow MgCl_2+H_2\)
\(2.............4............2..........2\)
\(V_{H_2}=2\cdot22.4=44.8\left(l\right)\)
\(m_{MgCl_2}=2\cdot95=190\left(g\right)\)
\(m_{\text{dung dịch sau phản ứng}}=2\cdot24+400-2\cdot2=444\left(g\right)\)
\(C\%MgCl_2=\dfrac{190}{444}\cdot100\%=42.79\%\)
a)
$Mg + 2HCl \to MgCl_2 + H_2$
b)
n HCl = 400.36,5%/36,5 = 4(mol)
n H2 = 1/2 n HCl = 2(mol)
V H2 = 2.22,4 = 44,8(lít)
c)
n MgCl2 = n H2 = 2(mol)
m MgCl2 = 2.95 = 190(gam)
d) n Mg = n H2 = 2(mol)
Sau phản ứng :
mdd = m Mg + mdd HCl - m H2 = 2.24 + 400 -2.2 = 444(gam)
C% MgCl2 = 190/444 .100% = 42,79%
Fe+H2SO4->FeSO4+H2
0,15---0,15-----0,15---0,15 mol
n Fe=8,4\56=0,15 mol
=>VH2=0,15.22,4=3,36l
=>m H2SO4=0,15.98=14,7g
=>C% H2SO4=14,7\245 .100=6%
=>m dd muối=8,4+245-0,15.2=253,1g
=>C% muối =0,15.152\253,1 .100=9%
\(n_{Zn}=\dfrac{13}{65}=0,2\left(mol\right)\\ n_{HCl}=\dfrac{109,5.20\%}{36,5}=0,6\left(mol\right)\\ PTHH:Zn+2HCl\rightarrow ZnCl_2+H_2\\ Vì:\dfrac{0,2}{1}< \dfrac{0,6}{2}\Rightarrow HCldư\\ n_{H_2}=n_{ZnCl_2}=n_{Zn}=0,2\left(mol\right)\\ n_{HCl\left(dư\right)}=0,6-0,2.2=0,2\left(mol\right)\\ a,V_{H_2\left(đktc\right)}=0,2.22,4=4,48\left(l\right)\\ b,m_{ZnCl_2}=136.0,2=27,2\left(g\right)\\ c,m_{ddsau}=13+109,5-0,2.2=122,1\left(g\right)\\ C\%_{ddZnCl_2}=\dfrac{27,2}{122,1}.100\approx22,277\%\\ C\%_{ddHCl\left(dư\right)}=\dfrac{0,2.36,5}{122,1}.100\approx5,979\%\)
Zn + 2HCl -> ZnCl2 + H2
a, nZn = 13/65= 0,2(mol)
mHCl= 109,5.20%/100%=21.9(g)
nHCl=21,9/36,5=0,6(mol)
Theo PT nHCl = 2nZn= 2.0,2= 0,4(mol)<0,6(mol)
=> HCl pư dư, Zn pư hết
Theo PT: nH2= nZn =0,2(mol)
VH2=0,2.22,4=4,48(l)
b, Theo PT: nZnCl2=nZn=0,2(mol)
mZnCl2= 0,2.136=27,2(g)
c, mdd sau pư= 13+109,5-0,2.2=122,1(g)
C%dd ZnCl2=27,2.100%/122,1=22,28%
nHCl dư= 0,6-0,4=0,2(mol)
mHcl
\(a)n_{Al}=\dfrac{10,8}{27}=0,4\left(mol\right)\\ 2Al+6HCl\xrightarrow[]{}2AlCl_3+3H_2\\ n_{H_2}=\dfrac{3}{2}n_{Al}=\dfrac{3}{2}\cdot0,4=0,6\left(mol\right)\\ V_{H_2}=0,6.22,4=13,44\left(l\right)\\ b)n_{HCl}=3n_{Al}=3.0,4=1,2\left(mol\right)\\ m_{HCl}=1,2.36,5=43,8\left(g\right)\\ m_{dd_{HCl}}=\dfrac{43,8}{10,95\%}\cdot100\%=400\left(g\right)\\ c)n_{AlCl_3}=n_{Al}=0,4mol\\ m_{AlCl_3}=0,4.133,5=53,4\left(g\right)\\ m_{H_2}=0,6.2=1,2\left(g\right)\\ m_{dd_{AlCl_3}}=10,8+400-1,2=409,6\left(g\right)\\ C_{\%AlCl_3}=\dfrac{53,4}{409,6}\cdot100\%\approx13\%\)
Câu 1:
a, \(Zn+2HCl\rightarrow ZnCl_2+H_2\)
b, \(n_{Zn}=\dfrac{16,25}{65}=0,25\left(mol\right)\)
\(n_{H_2}=n_{Zn}=0,25\left(mol\right)\)
\(\Rightarrow V_{H_2}=0,25.24,79=6,1975\left(l\right)\)
c, \(n_{HCl}=2n_{Zn}=0,5\left(mol\right)\)
\(\Rightarrow m_{HCl}=0,5.36,5=18,25\left(g\right)\)
\(\Rightarrow m_{ddHCl}=\dfrac{18,25}{10\%}=182,5\left(g\right)\)
d, \(n_{ZnCl_2}=n_{Zn}=0,25\left(mol\right)\)
\(\Rightarrow C\%_{ZnCl_2}=\dfrac{0,25.136}{16,25+182,5-0,25.2}.100\%\approx17,15\%\)
Câu 2:
a, \(2NaOH+H_2SO_4\rightarrow Na_2SO_4+2H_2O\)
c, \(n_{NaOH}=\dfrac{40}{40}=1\left(mol\right)\)
Theo PT: \(n_{H_2SO_4}=\dfrac{1}{2}n_{NaOH}=0,5\left(mol\right)\)
\(\Rightarrow V_{H_2SO_4}=\dfrac{0,5}{2}=0,25\left(l\right)\)
d, \(n_{Na_2SO_4}=\dfrac{1}{2}n_{NaOH}=0,5\left(mol\right)\)
\(\Rightarrow C_{M_{Na_2SO_4}}=\dfrac{0,5}{0,25}=2\left(M\right)\)
\(n_{Fe}=\dfrac{11,2}{56}=0,2\left(mol\right)\\ a,Fe+2HCl\rightarrow FeCl_2+H_2\\ n_{HCl}=2.0,2=0,4\left(mol\right)\\ b,m_{ddHCl}=\dfrac{0,4.36,5.100}{20}=73\left(g\right)\\ c,n_{H_2}=n_{FeCl_2}=n_{Fe}=0,2\left(mol\right)\\ m_{ddsau}=11,2+73-0,2.2=83,8\left(g\right)\\ C\%_{ddFeCl_2}=\dfrac{0,2.127}{83,8}.100\approx30,31\%\)