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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
\(VH_2=0,05.22,4=1,12\left(lít\right)\)
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, 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: \(n_{ZnCl_2}=n_{Zn}=0,5\left(mol\right)\)
\(\Rightarrow m_{ZnCl_2}=0,5.136=68\left(g\right)\)
a, \(CaCO_3+2HCl\rightarrow CaCl_2+CO_2+H_2O\)
b, \(n_{CaCO_3}=\dfrac{10}{100}=0,1\left(mol\right)\Rightarrow n_{CO_2}=n_{CaCO_3}=0,1\left(mol\right)\)
\(\Rightarrow V_{CO_2}=0,1.22,4=2,24\left(l\right)\)
c, \(n_{HCl}=2n_{CaCO_3}=0,2\left(mol\right)\Rightarrow C_{M_{HCl}}=\dfrac{0,2}{0,1}=2\left(M\right)\)
d, \(m_{NaOH}=550.10\%=55\left(g\right)\Rightarrow n_{NaOH}=\dfrac{55}{40}=1,375\left(mol\right)\)
\(\Rightarrow\dfrac{n_{NaOH}}{n_{CO_2}}=\dfrac{1,375}{0,1}=13,75>2\)
→ Pư tạo muối trung hòa Na2CO3.
PT: \(CO_2+2NaOH\rightarrow Na_2CO_3+H_2O\)
\(n_{Na_2CO_3}=n_{CO_2}=0,1\left(mol\right)\Rightarrow m_{Na_2CO_3}=0,1.106=10,6\left(g\right)\)
\(n_{Mg}=\dfrac{4,8}{24}=0,2\left(mol\right)\\ a,Mg+2HCl\rightarrow MgCl_2+H_2\\ n_{MgCl_2}=n_{H_2}=n_{Mg}=0,2\left(mol\right);n_{HCl}=0,2.2=0,4\left(mol\right)\\ b,C_{MddHCl}=\dfrac{0,4}{0,1}=4\left(M\right)\\ V_{H_2\left(đktc\right)}=0,2.22,4=4,48\left(l\right)\)
a)\(n_{HCl}=\dfrac{10,65}{36,5}=0,3mol\)
\(Fe+2HCl\rightarrow FeCl_2+H_2\)
0,15 0,3 0,15 0,15
b)\(V_{H_2}=0,15\cdot22,4=3,36l\)
c)\(n_{CuO}=\dfrac{16}{80}=0,2mol\)
\(CuO+H_2\rightarrow Cu+H_2O\)
0,2 0,15 0,15
\(m_{Cu}=0,15\cdot64=9,6g\)
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)`