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`Zn + 2HCl -> ZnCl_2 + H_2`
`0,2` `0,4` `0,2` `0,2` `(mol)`
`n_[Zn]=13/65=0,2(mol)`
`a)V_[H_2]=0,2.22,4=4,48(l)`
`b)C%_[HCl]=[0,4.36,5]/100 . 100 =14,6%`
`c)C%_[ZnCl_2]=[0,2.136]/[13+100-0,2.2].100~~24,16%`
`d)`
`H_2 + CuO` $\xrightarrow{t^o}$ `Cu + H_2 O`
`0,1` `0,1` `0,1` `(mol)`
`n_[CuO]=8/80=0,1(mol)`
Ta có:`[0,2]/1 > [0,1]/1`
`=>H_2` dư, `CuO` hết
`=>m_[Cu]=0,1.64=6,4(g)`
\(n_{Zn}=\dfrac{13}{65}=0,2mol\)
\(Zn+H_2SO_4\rightarrow ZnSO_4+H_2\)
0,2 0,2 0,2 0,2
a)\(V_{H_2}=0,2\cdot22,4=4,48l\)
b)\(m_{H_2SO_4}=0,2\cdot98=19,6g\)
\(C\%=\dfrac{m_{ct}}{m_{dd}}\cdot100\%=\dfrac{19,6}{100}\cdot100\%=19,6\%\)
c)\(m_{ZnSO_4}=0,2\cdot161=32,2g\)
\(m_{ddZnSO_4}=13+100-0,2\cdot2=112,6g\)
\(C\%=\dfrac{m_{ct}}{m_{dd}}\cdot100\%=\dfrac{32,2}{112,6}\cdot100\%=28,6\%\)
d)\(n_{CuO}=\dfrac{8}{80}=0,1mol\)
\(CuO+H_2\rightarrow Cu+H_2O\)
0,1 0,2 0,1
\(m_{Cu}=0,1\cdot64=6,4g\)
`Fe + 2HCl -> FeCl_2 + H_2 \uparrow`
`0,1` `0,2` `0,1` `0,1` `(mol)`
`n_[Fe]=[5,6]/56=0,1(mol)`
`a)V_[H_2]=0,1.22,4=2,24(l)`
`b)C_[M_[HCl]]=[0,2]/[0,1]=2(M)`
nHCl = 0.3*0.75 = 0.225 (mol)
Zn + 2HCl => ZnCl2 + H2
0.1125..0.225....0.1125..0.1125
mZn = 0.1125*65 = 7.3125 (g)
VH2 = 0.1125*22.4 = 2.52 (l)
CM ZnCl2 = 0.1125/0.3 = 0.375 (M)
nHCl= 0,75 x 0,3= 0,225(mol)
PTHH: Zn + 2HCl -> ZnCl2 + H2
a) 0,1125____0,225___0,1125___0,1125(mol)
=> a=mZn=0,1125.65=7,3125(g)
V(H2,đktc)=0,1125 x 22,4=2,52(l)
b) VddZnCl2=VddHCl=0,3(l)
=>CMddZnCl2=0,1125/0,3=0,375(M)
\(n_{Zn}=\dfrac{13}{65}=0,2\left(mol\right)\)
\(PTHH:Zn+H_2SO_4\rightarrow ZnSO_4+H_2\)
(mol)_____0,2____0,2______0,2____0,2__
\(a.V_{H_2}=22,4.0,2=4,48\left(l\right)\)
\(b.m_{ddH_2SO_4}=\dfrac{0,2.98.100}{24,5}=80\left(g\right)\)
\(c.m_{ddspu}=13+80-0,2.2=92,6\left(g\right)\\ \Rightarrow C\%_{ddspu}=\dfrac{0,2.136}{92,6}.100=29,4\left(\%\right)\)
a) Zn + 2HCl \(\rightarrow\) ZnCl2 + H2 (1)
nZn = 13/65 = 0,2(mol)
Theo PT(1) => nH2 = nZn = 0,2(mol)
=> VH2 = 0,2 . 22,4 = 4,48(l)
b) Theo PT(1) => nHCl = 2 . nZn = 2 . 0,2 = 0,4 (mol)
=> Vdd HCl = n : CM = 0,4 : 2 = 0,2(l)
c) Zn + H2SO4 \(\rightarrow\) ZnSO4 + H2 (2)
Theo PT(2) => nH2SO4 = nZn = 0,2(mol)
=> mH2SO4 = 0,2 . 98 = 19,6(g)
=> m dd H2SO4 24,5% = \(\dfrac{m_{ct}.100\%}{C\%}=\dfrac{19,6.100\%}{24,5\%}=80\left(g\right)\)