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`a)PTHH:`
`Mg + H_2 SO_4 -> MgSO_4 + H_2`
`0,2` `0,2` `0,2` `(mol)`
`n_[Mg]=[4,8]/24=0,2(mol)`
`b)m_[MgSO_4]=0,2.120=24(g)`
`c)C%_[MgSO_4]=24/[4,8+50-0,2.2].100~~44,12%`
a,Fe + 2HCl → FeCl + H2 (1)
FeO + 2HCl → FeCl + H2O (2)
nH2 = 3,36/ 22,4 = 0,15 ( mol)
Theo (1) nH2 = nFe = 0,15 ( mol)
mFe = 0,15 x 56 = 8.4 (g)
m FeO = 12 - 8,4 = 3,6 (g)
a, \(n_{H_2}=\frac{3,36}{22,4}=0,15\left(mol\right)\)
\(Fe+2HCl->FeCl_2+H_2\left(1\right)\)
\(FeO+2HCl->FeCl_2+H_2O\left(2\right)\)
theo (1) \(n_{Fe}=n_{H_2}=0,15\left(mol\right)\)
=> \(m_{Fe}=0,15.56=8,4\left(g\right)\)
=> \(m_{FeO}=12-8,4=3,6\left(g\right)\)
ta thấy : nFe =nH2 = 0,15
=> mFe =0,15 x 56 = 8,4g
%Fe=8,4/12 x 100 = 70%
=>%FeO = 100 - 70 = 30%
b) BTKLra mdd tìm mct of HCl
c) tìm mdd sau pứ -mH2 nha bạn
a)
\(n_{Zn}=\dfrac{13}{65}=0,2\left(mol\right)\)
PTHH: Zn + 2HCl --> ZnCl2 + H2
0,2-->0,4----->0,2--->0,2
=> VH2 = 0,2.22,4 = 4,48 (l)
b) mHCl = 0,4.36,5 = 14,6 (g)
=> \(m_{dd.HCl}=\dfrac{14,6.100}{7,3}=200\left(g\right)\)
c)
mdd sau pư = 13 + 200 - 0,2.2 = 212,6 (g)
mZnCl2 = 0,2.136 = 27,2 (g)
=> \(C\%=\dfrac{27,2}{212,6}.100\%=12,8\%\)
a) $2Al + 6HCl \to 2AlCl_3 + 3H_2$
b) n Al = 8,1/27 = 0,3(mol)
Theo PTHH :
n H2 = 3/2 n Al = 0,45(mol)
V H2 = 0,45.22,4 = 10,08(lít)
c) n AlCl3 = n Al = 0,3(mol)
m AlCl3 = 0,3.133,5 = 40,05(gam)
d) n HCl = 3n Al = 0,9(mol)
m dd HCl = 0,9.36,5/7,3% = 450(gam)
Sau phản ứng :
m dd = 8,1 + 450 -0,45.2 = 457,2(gam)
C% AlCl3 = 40,05/457,2 .100% = 8,76%
`a)PTHH:`
`2Al + 6HCl -> 2AlCl_3 + 3H_2`
`0,2` `0,6` `0,3` `(mol)`
`n_[Al]=[5,4]/27=0,2(mol)`
`b)V_[H_2]=0,3.22,4=6,72(l)`
`c)m_[dd HCl]=[0,6.36,5]/10 . 100 =219(g)`
a) \(Mg+H_2SO_4\rightarrow MgSO_4+H_2\)
\(2Al+3H_2SO_4\rightarrow Al_2\left(SO_4\right)_3+3H_2\)
b) \(n_{H_2\left(1\right)}=n_{Mg}=\dfrac{4,8}{24}=0,2\left(mol\right)\)
\(n_{H_2\left(2\right)}=\dfrac{3}{2}n_{Al}=\dfrac{3}{2}.\dfrac{2,7}{27}=0,15\left(mol\right)\)
=> \(V=\left(0,2+0,15\right).22,4=7,84\left(l\right)\)
c) \(n_{H_2SO_4\left(1\right)}=n_{Mg}=0,2\left(mol\right)\)
\(n_{H_2SO_4\left(2\right)}=\dfrac{3}{2}n_{Al}=0,15\left(mol\right)\)
=> \(m_{ddH_2SO_4}=\dfrac{0,35.98}{20\%}=171,5\left(g\right)\)
d) \(m_{ddsaupu}=4,8+2,7+171,5-0,35.2=178,3\left(g\right)\)
\(C\%_{MgSO_4}=\dfrac{120.0,1}{178,3}.100=6,73\%\)
\(C\%_{Al_2\left(SO_4\right)_3}=\dfrac{342.0,05}{178,3}.100=9,59\%\)
a,Mg+H2SO4-> MgSO4 +H2
2Al +3H2SO4 -> Al2(SO4)3 +3H2
b, n(Mg)=0,2mol
n(Al)=0,1mol
Số mol H2SO4=số mol H2= 0,2+ 0,1*3/2 =0,35mol
V(H2)= 7,84lit
c, MgSO4: m=0,2*120=24(g)
Al2(SO4)3 : m=342*0,05= 17,1(g)
d, khối lượng H2SO4= 0,35*98=34,3(g)
Khối lượng dd H2SO4 là:
m(dd)=34,3*100/20 = 171,5(g)
e,khối lượng dd sau pứ
m= m(Mg) +m(Al) + m(dd H2SO4) -m(H2) = 4,8+2,7+171,5-0,35*2=178,3(g)
C%(MgSO4)= 24*100%/178,3 =13,46%
C%(Al2SO4)3 = 17,1*100%/178,3 =9,59%