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\(n_{CuO}=\dfrac{2,4}{80}=0,03\left(mol\right)\\ PTHH:CuO+H_2\rightarrow\left(t^o\right)Cu+H_2O\\ n_{H_2}=n_{Cu}=n_{CuO}=0,03\left(mol\right)\\ a,V_{H_2\left(đktc\right)}=0,03.22,4=0,672\left(l\right)\\ b,m_{Cu}=0,03.64=1,92\left(g\right)\)
CuO+H2-to>Cu+H2O
0,06---0,06---0,06---0,06
n CuO=\(\dfrac{4,8}{80}=0,06mol\)
=>m Cu=0,06.64=3,84g
=>VH2=0,06.22,4=1,344l
c)Fe+2HCl->FeCl2+H2
0,06--------------------0,06 mol
=>m Fe=0,06.56=3,36g
nCuO = 4,8/80 = 0,06 (mol)
PTHH: CuO + H2 -> (t°) Cu + H2O
Mol: 0,06 ---> 0,06 ---> 0,06
mCu = 0,06.64 =3,84 (g)
VH2 = 0,06 . 22,4 = 1,344 (l)
PTHH: Fe + 2HCl -> FeCl2 + H2
Mol: 0,06 <--- 0,12 <--- 0,06 <--- 0,06
mFe = 0,06 . 56 = 3,36 (g)
mHCl = 0,12 . 36,5 = 4,38 (g)
PTHH : \(CuO+H_2-t^o->Cu+H_2O\)
----------0,03-----0,03------------0,03----0,03 ( mol )
Ta có nCuO= 2,4/80 =0,03
theo PTHH => nH2 =0,03
=> VH2 =0,03.22,4=0,672 lít
b) \(2Al+3H_2SO_4->Al_2\left(SO_4\right)_3+3H_2\)
nH2 =0,03 mol.Theo PTHH => nAl =0,02 mol
=> mAl =0,02.27=0,54 gam
=
PTHH: CuO + H2 -to-> Cu + H2O (1)
Ta có: \(n_{CuO}=\dfrac{2,4}{80}=0,03\left(mol\right)\)
a, \(n_{H_2}=n_{CuO}=0,03\left(mol\right)\\ V_{H_2\left(đktc\right)}=0,03.22,4=0,672\left(l\right)\)
b, PTHH: 2Al + 3H2SO4 -> Al2(SO4)3 + 3H2 (2)
Ta có: \(n_{H_2\left(2\right)}=n_{H_2\left(1\right)}=0,03\left(mol\right)\\ =>n_{Al\left(2\right)}=\dfrac{2.0,03}{3}=0,02\left(mol\right)\\ =>m_{Al}=0,02.27=0,54\left(g\right)\)
Bài 1:
nCuO = \(\dfrac{2,4}{80}=0,03\) mol
Pt: CuO + H2 --to--> Cu + H2O
0,03 mol-> 0,03 mol
Pt: 2Al + 3H2SO4 --> Al2(SO4)3 + 3H2
0,02 mol<------------------------------0,03 mol
mAl cần dùng = 0,02 . 27 = 0,54 (g)
\(a,n_{Mg}=\dfrac{2,4}{24}=0,1\left(mol\right)\\ PTHH:Mg+H_2SO_4\rightarrow MgSO_4+H_2\uparrow\left(1\right)\\ Theo.pt\left(1\right):n_{H_2}=n_{MgSO_4}=n_{Mg}=0,1\left(mol\right)\\ V_{H_2}=0,1.22,4=2,24\left(l\right)\\ m_{MgSO_4}=0,1.120=12\left(g\right)\\ b,PTHH:CuO+H_2\underrightarrow{t^o}Cu+H_2O\left(2\right)\\ Theo.pt\left(2\right):n_{Cu}=n_{H_2}=0,1\left(mol\right)\\ m_{Cu}=0,1.64=6,4\left(g\right)\)
1.\(n_{Zn}=\dfrac{13}{65}=0,2mol\)
\(Zn+2HCl\rightarrow ZnCl_2+H_2\)
0,2 0,2 ( mol )
\(V_{H_2}=0,2.22,4=4,48l\)
2.\(n_{CuO}=\dfrac{12}{80}=0,15mol\)
\(CuO+H_2\rightarrow\left(t^o\right)Cu+H_2O\)
0,15 < 0,2 ( mol )
0,15 0,15 ( mol )
\(m_{Cu}=0,15.64=9,6g\)
CuO+H2-to>Cu+H2O
0,03--0,03----0,03---0,03
n CuO=\(\dfrac{2,4}{80}\)=0,03 mol
=>VH2=0,03.22,4=0,672l
2Al+3H2SO4->Al2(SO4)3+3H2
0,02--------------------------------0,03
=>m Al=0,02.27=0,54g