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a) 2Al+ 3H2SO4-> Al2(SO4)3+3H2
b) 2Fe(OH)3+ 3H2SO4-> Fe2(SO4)3+ 6H2O
a) 2Al + 3H2SO4 \(\rightarrow\) Al2(SO4)3 + 3H2
b) 2Fe(OH)3 + 3H2SO4 \(\rightarrow\) Fe2(SO4)3 + 6H2O
a) PTHH :\(Fe_2O_3+3H_2SO_4->Fe_2\left(SO_4\right)_3+3H_2O\)
b) Ta có :\(n_{Fe_2O_3}=\dfrac{48}{160}=0,3\left(mol\right);n_{H_2SO_4}=\dfrac{58,8}{98}=0,6\left(mol\right)\)
\(Fe_2O_3+3H_2SO_4->Fe_2\left(SO_4\right)_3+3H_2O\)
1______3
\(\dfrac{1}{0,3}< \dfrac{3}{0,6}\)
=>\(n_{H_2SO_4}dư=>n_{H_2SO_{4\left(dư\right)}}=0,9mol=>m_{H_2SO_4\left(dư\right)}=0,9.98=88,2\left(g\right)\)
=>
a. 4P + 5O2 \(\underrightarrow{t^o}\) 2P2O5
b. 4Al + 3O2 \(\underrightarrow{t^o}\) 2Al2O3
c. Fe + 2HCl \(\rightarrow\) FeCl2 + H2\(\uparrow\)
d. H2 + CuO \(\underrightarrow{t^o}\) Cu + H2O
e. 3CO + Fe2O3 \(\underrightarrow{t^o}\) 2Fe + 3CO2\(\uparrow\)
f. Cu + 2H2SO4 \(\rightarrow\) CuSO4 + SO2\(\uparrow\) + 2H2O
g. Fe + 4HNO3 \(\rightarrow\) Fe(NO3)3 + NO\(\uparrow\) + 2H2O
h. 2Al + 3H2SO4 \(\rightarrow\) Al2(SO4)3 + 3H2\(\uparrow\)
i. Ca(HCO3)2 \(\underrightarrow{t^o}\) CaCO3 + CO2\(\uparrow\) + H2O
Bài 1:
\(4Al+3O_2\underrightarrow{t^o}2Al_2O_3\)
\(n_{Al_2O_3}=\dfrac{m}{M}=\dfrac{20,4}{102}=0,2\left(mol\right)\)
Theo PTHH, \(n_{Al}=2n_{Al_2O_3}=2\cdot0,2=0,4\left(mol\right)\)
\(m_{Al}=n\cdot M=0,4\cdot27=10,8\left(g\right)\)
Theo PTHH, \(n_{O_2}=\dfrac{3}{2}n_{Al_2O_3}=\dfrac{3}{2}\cdot0,2=0,3\left(mol\right)\)
\(V_{O_2}=n\cdot22,4=0,3\cdot22,4=6,72\left(l\right)\)
Bài 1:
4Al + 3O2 \(\underrightarrow{to}\) 2Al2O3
\(n_{Al_2O_3}=\dfrac{20,4}{102}=0,2\left(mol\right)\)
a) theo PT: \(n_{Al}=2n_{Al_2O_3}=2\times0,2=0,4\left(mol\right)\)
\(\Rightarrow m=m_{Al}=0,2\times27=5,4\left(g\right)\)
b) theo PT: \(n_{O_2}=\dfrac{3}{2}n_{Al_2O_3}=\dfrac{3}{2}\times0,2=0,3\left(mol\right)\)
\(\Rightarrow V_{O_2}=0,3\times22,4=6,72\left(l\right)\)
CuO + H2SO4 → CuSO4 + H2O (1)
Al2O3 + 3H2SO4 → Al2(SO4)3 + 3H2O (2)
Gọi \(x,y\) lần lượt là số mol của CuO và Al2O3
Theo PT1: \(n_{CuSO_4}=n_{CuO}=x\left(mol\right)\)
TheoPT2: \(n_{Al_2\left(SO_4\right)_3}=n_{Al_2O_3}=y\left(mol\right)\)
Ta có: \(\left\{{}\begin{matrix}160x+342y=57,9\\80x+102y=25,5\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}x=0,255\\y=0,05\end{matrix}\right.\)
Vậy \(n_{CuO}=0,255\left(mol\right);n_{Al_2O_3}=0,05\left(mol\right)\)
\(\Rightarrow m_{Al_2O_3}=0,05\times102=5,1\left(g\right)\)
\(\Rightarrow\%Al_2O_3=\dfrac{5,1}{25,5}\times100\%=20\%\)
a, Cu+2H2SO4(đ)-to-->CuSO4+SO2+2H2O
b, Zn+2H2SO4(đ)-to-->ZnSO4+SO2+2H2O
2Fe+6H2SO4(đ)-to-->Fe2(SO4)3+3SO2+6H2O
Cu + 2H2SO4 đặc → CuSO4 + 2H2O + SO2(to)
Ba + 2H2SO4 đặc → BaSO4 + 2H2O + SO2(to)
2Al + 6H2SO4 đặc → Al(SO4)3 + 6H2O + 3H2O (to)
pt: 2Al + 3H2SO4 => Al2(SO4)3 + 3H2
nAl = \(\dfrac{5,4}{27}=0,2mol\)
a) Theo pt: nH2 = \(\dfrac{3}{2}nAl=\dfrac{3}{2}.0,2=0,3mol\)
=> VH2 = 0,3.22,4 = 6,72 lít
b) Theo pt : nAl2(SO4)3 = \(\dfrac{1}{2}nAl=0,1mol\)
=> mAl2SO4 = 0,1.342 = 34,2 g