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nH2= 13,44 : 22,4 = 0,6 (mol)
pthh : 2Al + 3H2SO4 --> Al2(SO4)3 + 3H2
0,4 <------------------- 0,2<-----------<0,6 (mol)
mAl = 0,4 . 27 = 10,8 ( g)
mAl2(SO4)3= 0,2 . 342 = 68,4 (g)
nFe3O4 = 46,4 : 232 = 0,2 (mol)
pthh : Fe3O4 + 4H2 -t--> 3Fe + 4H2O
LTL :
0,2/1 > 0,6 /4
=> Fe3O4 du
theo pt nFe=3/4 nH2 = ,45 (mol)
=> mFe= 0,45 . 56= 25,2 (g)
\(a,n_{H_2}=\dfrac{13,44}{22,4}=0,6\left(mol\right)\\ PTHH:2Al+3H_2SO_4\rightarrow2Al_2\left(SO_4\right)_3+3H_2\uparrow\\ Theo.pt:n_{Al}=n_{Al_2\left(SO_4\right)_3}=\dfrac{2}{3}n_{H_2SO_4}=\dfrac{2}{3}.0,6=0,4\left(mol\right)\\ m_{Al}=0,4.27=10,8\left(g\right)\\ b,m_{Al_2\left(SO_4\right)_3}=0,4.342=136,8\left(g\right)\\ c,n_{Fe_3O_4}=\dfrac{46,4}{232}=0,2\left(mol\right)\\ PTHH:Fe_3O_4+4H_2\underrightarrow{t^o}3Fe+4H_2O\\ LTL:\dfrac{0,2}{1}>\dfrac{0,6}{4}\Rightarrow Fe_3O_4.du\\ n_{Fe}=\dfrac{3}{4}n_{H_2}=\dfrac{3}{4}.0,4=0,3\left(mol\right)\\ m_{Fe}=0,3.56=16,8\left(g\right)\)
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) nAl = \(\frac{40,5}{27}=1,5\left(mol\right)\)
Từ PT \(\Rightarrow n_{H_2SO_4}=2,25\left(mol\right);n_{Al_2\left(SO_4\right)_3}=0,75\left(mol\right);n_{H_2}=2,25\left(mol\right)\)
\(\Rightarrow m_{H_2SO_4}=2,25.98=220,5\left(g\right)\)
c) \(m_{Al_2\left(SO_4\right)_3}=0,75.342=256,5\left(g\right)\)
d) đktc : \(V_{H_2}=22,4.2,25=50,4\left(l\right)\)
a) 2Al + 3H2SO4 → Al2(SO4)3 + 3H2 (1)
b) nAl = 40,5 : 27 = 1,5 mol
Từ pt(1) suy ra : nH2SO4 = \(\frac{3}{2}nAl\) = \(\frac{3}{2}.1,5=2,25mol\)
Khối lượng H2SO4 là : mH2SO4 = 2,25 . 98 = 220,5 g
c) Từ pt(1) => nAl2(SO4)3 = \(\frac{1}{2}nAl=\frac{1}{2}.1,5=0,75mol\)
=> mAl2(SO4)3 = 0,75 . 342 = 256,5 g
d) Từ pt(1) => nH2 = nH2SO4 = 2,25 mol
Thể tích khí H2 là : VH2=2,25 . 22,4 = 50,4 lit
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 )
\(m_{Fe}=0,1.56=5,6g\)
`Fe + 2HCl -> FeCl_2 + H_2`
`0,2` `0,4` `0,2` `0,2` `(mol)`
`n_[Fe] = [ 11,2 ] / 56 = 0,2 (mol)`
`a) V_[H_2] = 0,2 . 22, 4= 4,48 (l)`
`b) m_[HCl] = 0,4 . 36,5 = 14,6 (g)`
`c) m_[FeCl_2] = 0,2 . 127 = 25,4 (g)`
pthh 4fe+ 6hcl -> 2fe2cl3+ 3h2
tính số mol của fe:.....................
tính thể tính khí h2 V=n.22,4= (l)
khối lượng hcl là m = n.M= (g)
khối lg fe2cl3 là m=n.M = (g)
chúc bạn học tốt:)))
a)
\(2Al + 3H_2SO_4 \to Al_2(SO_4)_3 + 3H_2\)
b)
\(n_{H_2} = \dfrac{6.13,44}{22,4} = 3,6(mol)\)
Theo PTHH :
\(n_{Al} = \dfrac{2}{3}n_{H_2} = 2,4(mol)\\ \Rightarrow m_{Al} = 2,4.27 = 64,8(gam)\)
c)
\(4Al + 3O_2 \xrightarrow{t^o} 2Al_2O_3\)
Theo PT trên :
\(n_{O_2} = \dfrac{3}{4}n_{Al} = 1,8(mol)\\ \Rightarrow V_{O_2} = 1,8.22,4 = 40,32(lít)\)
`Mg + 2HCl -> MgCl_2 + H_2`
`0,15` `0,3` `0,15` `(mol)`
`n_[Mg]=[3,6]/24=0,15(mol)`
`a)V_[H_2]=0,15.22,4=3,36(l)`
`b)m_[HCl]=0,3.36,5=10,95(g)`
`c)`
`H_2 + CuO` $\xrightarrow{t^o}$ `Cu + H_2 O`
`0,15` `0,15` `(mol)`
`=>m_[Cu]=0,15.64=9,6(g)`
\(n_{Mg}=\dfrac{3,6}{24}=0,15\left(mol\right)\)
PTHH: Mg + 2HCl ---> MgCl2 + H2
0,15->0,3------------------>0,15
CuO + H2 --to--> Cu + H2O
0,15------>0,15
=> \(V_{H_2}=0,15.22,4=3,36\left(l\right)\\ m_{HCl}=0,3.36,5=10,95\left(g\right)\\ m_{Cu}=0,15.64=9,6\left(g\right)\)
nH2SO4 = 14,7: 27=0,54(mol)
PTHH : 2Al + 3H2SO4 --> Al2(SO4)3 + 3H2
theo pt , nH2 = nH2SO4=0,54(mol)
=> VH2(đktc) = 0,54. 22,4=12,096 (l)
b theo pt nAl = 3/2. nH2=0,36 (mol)
=> mAl = 0,36.27 =9,72(g)
c)theo pt n Al2(SO4)3 = 1/2nAl = 0,18(mol)
=>mAl2(SO4)3= 0,18.342=61,56(g)