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a.b.
\(n_{Zn}=\dfrac{13}{65}=0,2mol\)
\(Zn+2HCl\rightarrow ZnCl_2+H_2\)
0,2 0,2 ( mol )
\(V_{H_2}=0,2.22,4=4,48l\)
c.
\(CuO+H_2\rightarrow\left(t^o\right)Cu+H_2O\)
0,3 > 0,2 ( mol )
0,2 0,2 ( mol )
\(m_{Cu}=0,2.64=12,8g\)
\(nFe=\dfrac{2,8}{56}=0,05\left(mol\right)\)
\(Fe+2HCl\rightarrow FeCl_2+H_2\)
0,05 0,05
\(VH_2=0,05.22,4=1,12\left(lít\right)\)
a, PT: \(Zn+2HCl\rightarrow ZnCl_2+H_2\)
b, Ta có: \(n_{Zn}=\dfrac{32,5}{65}=0,5\left(mol\right)\)
Theo PT: \(n_{H_2}=n_{Zn}=0,5\left(mol\right)\)
\(\Rightarrow V_{H_2}=0,5.22,4=11,2\left(l\right)\)
c, Theo PT: \(n_{ZnCl_2}=n_{Zn}=0,5\left(mol\right)\)
\(\Rightarrow m_{ZnCl_2}=0,5.136=68\left(g\right)\)
1) $Zn + 2HCl \to ZnCl_2 + H_2$
2) $n_{H_2} = n_{Zn} = \dfrac{13}{65} = 0,2(mol)$
$V_{H_2} = 0,2.22,4 = 4,48(lít)$
a. PTHH: Fe + 2HCl ---> FeCl2 + H2 (1)
b, \(n_{Fe}=\dfrac{11,2}{56}=0,2\left(mol\right)\)
Theo pthh (1): \(n_{H_2}=n_{Fe}=0,2\left(mol\right)\)
\(\rightarrow V_{H_2}=0,2.22,4=4,48\left(l\right)\)
c, PTHH: 2H2 + O2 --to--> 2H2O (2)
Theo pthh (2): \(n_{O_2}=\dfrac{1}{2}n_{H_2}=\dfrac{1}{2}.0,2=0,1\left(mol\right)\)
\(\rightarrow m_{O_2}=0,1.32=3,2\left(g\right)\)
\(n_{Mg}=\dfrac{4,8}{24}=0,2\left(mol\right)\\ a,Mg+2HCl\rightarrow MgCl_2+H_2\\ n_{MgCl_2}=n_{H_2}=n_{Mg}=0,2\left(mol\right);n_{HCl}=0,2.2=0,4\left(mol\right)\\ b,C_{MddHCl}=\dfrac{0,4}{0,1}=4\left(M\right)\\ V_{H_2\left(đktc\right)}=0,2.22,4=4,48\left(l\right)\)
\(n_{HCl}=0,3.1=0,3\left(mol\right)\\
pthh:Fe+2HCl\rightarrow FeCl_2+H_2\)
0,15 0,3 0,15
\(m_{Fe}=0,15.56=8,4\left(g\right)\\
V_{H_2}=0,15.22,4=3,36\left(l\right)\)
a, \(Zn+H_2SO_4\rightarrow ZnSO_4+H_2\)
b, \(n_{Zn}=\dfrac{6,5}{65}=0,1\left(mol\right)\)
\(n_{H_2SO_4}=\dfrac{19,6}{98}=0,2\left(mol\right)\)
Xét tỉ lệ: \(\dfrac{0,1}{1}< \dfrac{0,2}{1}\), ta được H2SO4 dư.
Theo PT: \(n_{H_2}=n_{Zn}=0,1\left(mol\right)\Rightarrow V_{H_2}=0,1.22,4=2,24\left(l\right)\)
`a)PTHH:`
`Zn + 2HCl -> ZnCl_2 + H_2`
`0,3` `0,3` `(mol)`
`n_[Zn]=[19,5]/65=0,3(mol)`
`b)V_[H_2]=0,3.22,4=6,72(l)`