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a.b.c.\(n_{Al}=\dfrac{2,7}{27}=0,1mol\)
\(2Al+6HCl\rightarrow2AlCl_3+3H_2\)
0,1 0,1 0,15 ( mol )
\(V_{H_2}=0,15.22,4=3,36l\)
\(m_{AlCl_3}=0,1.133,5=13,35g\)
d.\(Fe_2O_3+3H_2\rightarrow\left(t^o\right)2Fe+3H_2O\)
0,15 0,1 ( mol )
\(m_{Fe}=0,1.56=5,6g\)
\(a,n_{Al}=\dfrac{4,5}{27}=\dfrac{1}{6}\left(mol\right)\)
PTHH: \(2Al+3H_2SO_4\rightarrow Al_2\left(SO_4\right)_3+3H_2\)
\(\dfrac{1}{6}\)-->\(0,25\)-------->\(\dfrac{1}{12}\)------------>0,25
\(V_{ddH_2SO_4}=\dfrac{0,25}{1,5}=\dfrac{1}{6}\left(l\right)\\ b,m_{muối}=\dfrac{1}{12}.342=28,5\left(g\right)\\ V_{H_2}=0,25.22,4=5,6\left(l\right)\)
\(n_{Na}=\dfrac{2,3}{23}=0,1\left(mol\right)\)
PTHH: 2Na + 2H2O --> 2NaOH + H2
0,1-------------->0,1---->0,05
=> VH2 = 0,05.22,4 = 1,12 (l)
mdd sau pư = 2,3 + 197,8 - 0,05.2 = 200 (g)
=> \(C\%=\dfrac{0,1.40}{200}.100\%=2\%\)
\(V_{dd}=\dfrac{200}{1,08}=\dfrac{5000}{27}\left(ml\right)=\dfrac{5}{27}\left(l\right)\)
=> \(C_M=\dfrac{0,1}{\dfrac{5}{27}}=0,54M\)
Lớp 8 rồi ghi đề cho đúng đứng đắn vào:v
\(n_{H_2SO_4}=0,2.1=0,2\left(mol\right)\)
\(Fe+H_2SO_4\rightarrow FeSO_4+H_2\)
0,2<--- 0,2 ----> 0,2 -------> 0,2
\(\%_{m_{Fe}}=\dfrac{56.0,2.100\%}{15,8}=70,89\%\\ \Rightarrow\%_{m_{Cu}}=100\%-70,89\%=29,11\%\)
b
\(V_{H_2}=0,2.22,4=4,48\left(l\right)\)
c
\(FeSO_4+2NaOH\rightarrow Fe\left(OH\right)_2+Na_2SO_4\)
0,2 ------------------------> 0,2
\(m_{Fe\left(OH\right)_2}=0,2.90=18\left(g\right)\)
\(n_{H_2SO_4}=0,2.1=0,2\left(mol\right)\)
PTHH :
\(Fe+H_2SO_4\rightarrow FeSO_4+H_2\uparrow\)
0,2 0,2 0,2 0,2
\(m_{Fe}=0,2.56=11,2\left(g\right)\)
\(\%m_{Fe}=\dfrac{11,2}{15,8}.100\%\approx70,89\%\)
\(\%m_{Cu}=100\%-70,89\%\approx29,11\%\)
\(b,V_{H_2}=0,2.22,4=4,48\left(l\right)\)
\(c,FeSO_4+2NaOH\rightarrow Fe\left(OH\right)_2\downarrow+Na_2SO_4\)
0,2 0,2
\(m_{Fe\left(OH\right)_2}=0,2.90=18\left(g\right)\)
PTHH: \(2Al+3H_2SO_4\rightarrow Al_2\left(SO_4\right)_3+3H_2\)
Ta có: \(n_{Al}=\dfrac{2,7}{27}=0,1\left(mol\right)\)
\(\Rightarrow\left\{{}\begin{matrix}n_{H_2SO_4}=0,15\left(mol\right)=n_{H_2}\\n_{Al_2\left(SO_4\right)_3}=0,05\left(mol\right)\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}m_{H_2SO_4}=0,15\cdot98=14,7\left(g\right)\\m_{Al_2\left(SO_4\right)_3}=0,05\cdot342=17,1\left(g\right)\\V_{H_2}=0,15\cdot22,4=3,36\left(l\right)\end{matrix}\right.\)
a) \(n_{Al}=\dfrac{2,7}{27}=0,1\left(mol\right)\)
PTHH: 2Al + 3H2SO4 --> Al2(SO4)3 + 3H2
______0,1--->0,15-------->0,05------->0,15
=> mH2SO4 = 0,15.98 = 14,7 (g)
b) VH2 = 0,15.22,4 = 3,36 (l)
c) mAl2(SO4)3 = 0,05.342 = 17,1 (g)
a) 2Al + 3H2SO4 --> Al2(SO4)3 + 3H2
b) \(n_{Al}=\dfrac{10,8}{27}=0,4\left(mol\right)\)
PTHH: 2Al + 3H2SO4 --> Al2(SO4)3 + 3H2
0,4--->0,6-------------------->0,6
=> VH2 = 0,6.22,4 = 13,44 (l)
c) \(V_{dd.H_2SO_4}=\dfrac{0,6}{1}=0,6\left(l\right)\)
d) \(n_{Fe_2O_3}=\dfrac{16}{160}=0,1\left(mol\right)\)
PTHH: Fe2O3 + 3H2 --to--> 2Fe + 3H2O
Xét tỉ lệ: \(\dfrac{0,1}{1}< \dfrac{0,6}{3}\) => Fe2O3 hết, H2 dư
PTHH: Fe2O3 + 3H2 --to--> 2Fe + 3H2O
0,1----------------->0,2
=> mFe = 0,2.56 = 11,2 (g)
Câu 12: Cho 5,4g nhôm tác dụng với dung dịch HCl thì thể tích khí hiđro sinh ra ở đktc là bao nhiêu?
\(n_{Al}=\dfrac{5.4}{27}=0.2\left(mol\right)\)
\(2Al+6HCl\rightarrow2AlCl_3+3H_2\)
\(0.2....................................0.3\)
\(V_{H_2}=0.3\cdot22.4=6.72\left(l\right)\)
nAl=5,4/27=0,2(mol)
PTHH: 2Al + 6HCl -> 2AlCl3 +3 H2
Ta có: nH2=3/2. 0,2=0,3(mol)
=>V(H2,đktc)=0,3 x 22,4=6,72(l)
\(2Al+3H_2SO_4\rightarrow Al_2\left(SO_4\right)_3+3H_2\)
0,5\(\rightarrow\) 0,75 (MOL)
\(V_{H_2}=n_{H_2}.22,4=0,75.22,4=16,8\) (l)