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a)
\(n_{Zn}=\dfrac{13}{65}=0,2\left(mol\right)\)
PTHH: Zn + 2HCl --> ZnCl2 + H2
0,2-->0,4----->0,2--->0,2
=> VH2 = 0,2.22,4 = 4,48 (l)
b) mHCl = 0,4.36,5 = 14,6 (g)
=> \(m_{dd.HCl}=\dfrac{14,6.100}{7,3}=200\left(g\right)\)
c)
mdd sau pư = 13 + 200 - 0,2.2 = 212,6 (g)
mZnCl2 = 0,2.136 = 27,2 (g)
=> \(C\%=\dfrac{27,2}{212,6}.100\%=12,8\%\)
\(n_{Zn}=\dfrac{13}{65}=0,2\left(mol\right)\)
\(PTHH:Zn+H_2SO_4\rightarrow ZnSO_4+H_2\)
(mol)_____0,2____0,2______0,2____0,2__
\(a.V_{H_2}=22,4.0,2=4,48\left(l\right)\)
\(b.m_{ddH_2SO_4}=\dfrac{0,2.98.100}{24,5}=80\left(g\right)\)
\(c.m_{ddspu}=13+80-0,2.2=92,6\left(g\right)\\ \Rightarrow C\%_{ddspu}=\dfrac{0,2.136}{92,6}.100=29,4\left(\%\right)\)
\(n_{Mg}=\dfrac{10,8}{24}=0,45\left(mol\right)\\
pthh:Mg+H_2SO_4\rightarrow MgSO_4+H_2\uparrow\)
0,45 0,45 0,45
\(m_{H_2SO_4}=0,45.98=44,1\left(g\right)\\
C\%_{H_2SO_4}=\dfrac{44,1}{176,4}.100\%=25\%\\
V_{H_2}=0,45.22,4=10,08\left(l\right)\)
\(n_{Zn}=\dfrac{13}{65}=0,2\left(mol\right)\)
PTHH: Zn + H2SO4 ---> ZnSO4 + H2
0,2--->0,2--------->0,2------>0,2
\(V_{H_2}=0,2.22,4=4,48\left(l\right)\\ m_{H_2SO_4}=\dfrac{0,2.98}{20\%}=98\left(g\right)\\ \rightarrow V_{ddH_2SO_4}=\dfrac{98}{1,14}=86\left(ml\right)=0,086\left(l\right)\\ \rightarrow C_{M\left(H_2SO_4\right)}=\dfrac{0,2}{0,086}=2,33M\)
a) \(n_{Fe}=\dfrac{1,12}{56}=0,02\left(mol\right)\)
PTHH: Fe + H2SO4 ---> FeSO4 + H2
0,02->0,02------------------>0,02
b) \(V_{H_2}=0,02.22,4=0,448\left(l\right)\)
c) \(C\%_{H_2SO_4}=\dfrac{0,02.98}{100}.100\%=1,96\%\)
d) \(n_{CuO}=\dfrac{0,8}{80}=0,01\left(mol\right)\)
PTHH: CuO + H2 --to--> Cu + H2O
LTL: 0,01 < 0,02 => H2 dư
Theo pthh: nCu = nCuO = 0,01 (mol)
=> mCu = 0,01.64 = 0,64 (g)
a)
$Fe + 2HCl \to FeCl_2 + H_2$
n H2 = n Fe = 11,2/56 = 0,2(mol)
V H2 = 0,2.22,4 = 4,48(lít)
b)
n HCl = 2n Fe = 0,2.2 = 0,4(mol)
=> CM HCl = 0,4/0,4 = 1M
c)
$CuO + H_2 \xrightarrow{t^o} Cu + H_2O$
Ta thấy :
n CuO = 64/80 = 0,8 > n H2 = 0,2 nên CuO dư
Theo PTHH :
n CuO pư = n Cu = n H2 = 0,2(mol)
n Cu dư = 0,8 - 0,2 = 0,6(mol)
Vậy :
%m Cu = 0,2.64/(0,2.64 + 0,6.80) .100% = 21,05%
%m CuO = 100% -21,05% = 78,95%
a, \(n_{Mg}=\dfrac{4,8}{24}=0,2\left(mol\right)\)
PTHH: Mg + 2HCl → MgCl2 + H2
Mol: 0,2 0,4 0,2 0,2
\(V_{H_2}=0,2.22,4=4,48\left(l\right)\)
b,\(C\%_{ddHCl}=\dfrac{0,4.36,5.100\%}{200}=7,3\%\)
c, mdd sau pứ = 4,8+200-0,2.2 = 204,4 (g)
\(C\%_{ddMgCl_2}=\dfrac{0,2.95.100\%}{204,4}=9,3\%\)
\(Mg+H_2SO_4\rightarrow MgSO_4+H_2\)
0,305 0,305 0,305 0,305
\(n_{H_2}=\dfrac{6,832}{22,4}=0,305\left(mol\right)\)
\(a,m_{H_2SO_4}=98.0,305=29,89\left(g\right)\)
\(m_{ddH_2SO_4}=\dfrac{28,89}{12,5}.100\approx199,3\left(g\right)\)
\(m_{Mg}=24.0,305=7,32\left(g\right)\)
\(m_{H_2}=0,305.2=0,61\left(g\right)\)
Áp dụng định luật bảo toàn khổi lượng , ta có :
\(m_{MgSO_4}=\left(199,3+7,32\right)-0,61=206,01\left(g\right)\)
\(b,m_{MgSO_4}=0,305.120=36,6\left(g\right)\)
\(C\%_{MgSO_4}=\dfrac{36,6}{206,01}.100\%\approx17,8\%\)