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\(n_{Al}=\dfrac{27}{27}=1\left(mol\right)\\ pthh:2Al+6HCl\rightarrow2AlCl_3+3H_2\)
1 3 1,5
\(V_{H_2}=1,5.22,4=33,6\left(l\right)\\
C_{M\left(HCl\right)}=\dfrac{3}{0,3}=10M\)
\(pthh:Fe_2O_3+3H_2\underrightarrow{t^o}2Fe+3H_2O\)
1,5 1
\(m_{Fe}=1.56=56g\)
\(n_{Zn}=\dfrac{26}{65}=0,4\left(mol\right)\\
pthh:Zn+2HCl\rightarrow ZnCl_2+H_2\)
\(LTL:\dfrac{0,4}{1}>\dfrac{0,5}{2}\)
→Zn dư
\(n_{Zn\left(p\text{ư}\right)}=\dfrac{1}{2}n_{HCl}=0,25\left(mol\right)\\
m_{Zn\left(d\right)}=\left(0,4-0,25\right).65=9,75g\)
![](https://rs.olm.vn/images/avt/0.png?1311)
4. a) PTHH: \(Zn+2HCl\rightarrow ZnCl_2+H_2\)
b) \(n_{Zn}=\dfrac{6,5}{65}=0,1\left(mol\right)\)
Theo PTHH: \(n_{H_2}=n_{Zn}=0,1\left(mol\right)\)
\(\Rightarrow V_{H_2}=0,1.22,4=2,24\left(l\right)\)
c) \(C_M=\dfrac{n_{ct}}{V_{HCl}}=\dfrac{0,1}{0,8}=0,125M\)
d) \(H_2+CuO\rightarrow Cu+H_2O\)
\(n_{CuO}=\dfrac{0,8}{80}=0,1\left(mol\right)\)
Ta có: \(\dfrac{n_{H_2}}{1}=\dfrac{n_{CuO}}{1}\)
Vậy không có chất nào dư
\(\Rightarrow n_{Cu}=n_{H_2}=0,1\left(mol\right)\)
\(\Rightarrow m_{Cu}=0,1.64=6,4\left(g\right)\)
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nS = 6,4/32 = 0,2 (mol)
nO2 = 2,24/22,4 = 0,1 (mol)
PTHH: S + O2 -> (t°) SO2
LTL: 0,2 > 0,1 => S dư
nS (p/ư) = nSO2 = nO2 = 0,1 (mol)
=> VSO2 = 0,1 . 22,4 = 2,24 (l)
=> mS (dư) = (0,2 - 0,1) . 32 = 3,2 (g)
![](https://rs.olm.vn/images/avt/0.png?1311)
PTHH: \(Mg\left(OH\right)_2\rightarrow^{t^o}MgO+H_2O\)
Vậy tỉ lệ là \(1:1:1\)
\(n_{CaO}=\dfrac{8,4}{56}=0,15\left(mol\right)\)
PTHH: CaCO3 --to--> CaO + CO2
0,15<-----------0,15
=> m = 0,15.100 = 15 (g)
Anh kudo giải hóa hoài z a