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`a)PTHH:`
`2Al + 6HCl -> 2AlCl_3 + 3H_2`
`0,2` `0,6` `0,3` `(mol)`
`n_[Al]=[5,4]/27=0,2(mol)`
`b)V_[H_2]=0,3.22,4=6,72(l)`
`c)m_[dd HCl]=[0,6.36,5]/10 . 100 =219(g)`
a)
\(Zn + 2HCl \to ZnCl_2 + H_2\)
b),c)
Theo PTHH :
\(n_{ZnCl_2} = n_{H_2} = n_{Zn} = \dfrac{13}{65} = 0,2(mol)\)
Vậy :
\(m_{ZnCl_2} = 0,2.136 = 27,2(gam)\\ V_{H_2} =0,2.22,4 = 4,48(lít)\)
a. Zn + 2HCl → ZnCl2 + H2
b. nZn = n\(_{ZnCl_2}\) =\(\dfrac{13}{65}=0,2\left(mol\right)\) => m\(_{ZnCl_2}\)= 0,2.136 = 27,2(g)
c. n\(_{H_2}\)= nZn = 0,2 (mol) => V\(_{H_2}\)=0,2.22,4 = 4,48 (lít)
\(a) Mg + 2HCl \to MgCl_2 + H_2\\ b) n_{MgCl_2} = n_{Mg} = \dfrac{0,24}{24} = 0,01(mol)\\ m_{MgCl_2} = 0,01.95 = 0,95(gam)\\ c) n_{H_2} = n_{Mg} = 0,01(mol) \Rightarrow V_{H_2} = 0,01.22,4 = 0,224(lít)\)
PTHH: \(2Al+3H_2SO_4\rightarrow Al_2\left(SO_4\right)_3+3H_2\uparrow\)
Làm gộp các phần còn lại
Ta có: \(n_{Al}=\dfrac{5,4}{27}=0,2\left(mol\right)\) \(\Rightarrow\left\{{}\begin{matrix}n_{Al_2\left(SO_4\right)_3}=0,1mol\\n_{H_2SO_4}=n_{H_2}=0,3mol\end{matrix}\right.\)
\(\Rightarrow\left\{{}\begin{matrix}V_{H_2}=0,3\cdot22,4=6,72\left(l\right)\\m_{Al_2\left(SO_4\right)_3}=0,1\cdot342=34,2\left(g\right)\\m_{H_2SO_4}=0,3\cdot98=29,4\left(g\right)\end{matrix}\right.\)
2Al+6HCl->2AlCl3+3H2
0,4--------------------------0,6
n Al=0,4 mol
=>VH2=0,6.22,4=13,44l
H2+HgO-tO>Hg+H2O
0,6--------------0,6
=>m Hg=0,6.201=120,6g
\(n_{Al}=\dfrac{10,8}{27}=0,4\left(mol\right)\\ PTHH:2Al+6HCl\rightarrow2AlCl_3+3H_2\)
0,4 0,6
\(\rightarrow V_{H_2}=0,6.22,4=13,44\left(l\right)\\ PTHH:HgO+H_2\underrightarrow{t^o}Hg+H_2P\)
0,6 0,6
\(\rightarrow m_{Hg}=0,6.201=120,6\left(g\right)\)
\(n_{AlCl_3}=\dfrac{26.7}{133.5}=0.2\left(mol\right)\)
\(2Al+6HCl\rightarrow2AlCl_3+3H_2\)
\(.........0.6......0.2.......0.3\)
\(V_{H_2}=0.3\cdot22.4=6.72\left(l\right)\)
\(C\%HCl=\dfrac{0.6\cdot36.5}{300}\cdot100\%=7.3\%\)