a) nFe= 0,25(mol)

PTHH: Fe + H2SO4 -> FeSO4 + H2

0,25______0,25______0,25__0,25(mol)

b) V(H2,đktc)=0,25.22,4=5,6(l)

c) mH2SO4= 0,25.98= 24,5(g)

Lời giải "chi tiết"

PTHH: \(Fe+H_2SO_4\rightarrow FeSO_4+H_2\uparrow\)

Ta có: \(n_{Fe}=\dfrac{14}{56}=0,25\left(mol\right)=n_{H_2SO_4}=n_{H_2}\)

\(\Rightarrow\left\{{}\begin{matrix}m_{H_2SO_4}=0,25\cdot98=24,5\left(g\right)\\V_{H_2}=0,25\cdot22,4=5,6\left(l\right)\end{matrix}\right.\)

a,PTHH:

=0,25(mol)

Theo pt: nH2=nFe=0,25(mol)

⇒VH2=0,25.22,4=5,6(l)

c, Theo pt: nH2SO4=nFe=0,25(mol)

⇒mH2SO4=0,25.98=24,5(g)

PTHH: Zn + 2 HCl -> ZnCl2 + H2

nZn=13/65=0,2(mol)

nH2=nZnCl2=nZn=0,2(mol)

a) V(H2,đktc)=0,2.22,4=4,48(l)

b) nCuO= 24/80=0,3(mol)

PTHH: CuO + H2 -to-> Cu + H2O

Ta có: 0,3/1 > 0,2/1 => CuO dư, H2 hết, tính theo nH2

=> nCu=nH2=0,2(mol)

=> mCu=0,2.64=12,8(g)

c) 2 H2 + O2 -to-> H2O

nO2= 1/2 . nH2= 1/2 . 0,2=0,1(mol)

=> m(O2,đktc)=0,1.22,4=2,24(l)

nZn=6,5/65=0,1mol

đổi 500ml = 0,5l

nHCl = 0,5.1 =0,5mol

pt : Zn + 2HCl ------> ZnCl2 + H2

n có: 0,1 0,5

n pứ: 0,1 ----->0,2----------->0,1-------->0,1

n dư: 0 0,3

VH2= 0,1.22.4 =2,24l

mZnCl2= 0,1.136=13,6g

Vdd sau pứ =VddHCl =0,5l

CM(HCl dư) = 0,3/0,5=0,6M

CM(ZnCl2) = 0,1/0,5 =0,2M

a)

n_Zn = \(\dfrac{6,5}{65}=0,1mol\)

500ml = 0,5 (l)

Ta có: C_M = \(\dfrac{n}{V}=\dfrac{n}{0,5}=1\Rightarrow n_{HCl}=0,5mol\)

Zn + 2HCl -----> ZnCl₂ + H₂

0,1 0,5

Xét: \(\dfrac{0,1}{1}< \dfrac{0,5}{2}\Rightarrow\) Zn hết, HCl dư

\(\Rightarrow\) \(n_{H_2}=0,1mol\) \(\Rightarrow V_{H_2}=0,1.22,4=2,24l\)

b) \(n_{ZnCl_2}=0,1mol\)

\(\Rightarrow\)\(m_{ZnCl_2}=0,1.136=13,6g\)

c) C_M = \(\dfrac{n_{ZnCl_2}}{V}\) = \(\dfrac{0,1}{0,5}\)= 0,2M

n_{HCl dư} = 0,5 - 0,2 = 0,3 mol

C_M = \(\dfrac{n_{HCL}}{V}\) = \(\dfrac{0,3}{0,5}\) = 0,6M

nNa=11,5/23=0,5(mol)

nH2O=189/18=10,5(mol)

2Na+2H2O--->2NaOH+H2

2_____2

0,5___10,5

Ta có: 0,5/2<10,5/2

=>H2O dư

Theo pt: nH2=1/2nNa=1/2.0,5=0,25(mol)

=>VH2=0,25.22,4=5,6(l)

Theo pt: nNaOH=nNa=0,5(mol)

=>mNaOH=0,5.40=20(g)

mdd=11,5+189-0,25.2=200(g)

=>C%=20/200.100%=10%

nFeO=14,4/72=0,2(mol)

FeO+H2--->Fe+H2O

1____1

0,2___0,25

Ta có: 0,2/1<0,25/1

=>H2 dư

Theo pt: nFe=nFeO=0,2(mol)

=>mFe=0,2.56=11,2(g)

===>mFe thu đc= 11,2.75%=8,4(g)

Fe + H₂SO₄ → FeSO₄ + H₂ (1)

\(n_{Fe}=\frac{5,6}{56}=0,1\left(mol\right)\)

\(n_{H_2SO_4}=0,5\times0,2=0,1\left(mol\right)\)

Theo PT1: \(n_{Fe}=n_{H_2SO_4}\)

Theo bài: \(n_{Fe}=n_{H_2SO_4}\)

Vì \(1=1\) ⇒ phản ứng vừa đủ

a) Theo PT1: \(n_{H_2}=n_{Fe}=0,1\left(mol\right)\)

\(\Rightarrow V_{H_2}=0,1\times22,4=2,24\left(l\right)\)

b) Theo PT1: \(n_{FeSO_4}=n_{Fe}=0,1\left(mol\right)\)

\(\Rightarrow C_{M_{FeSO_4}}=\frac{0,1}{0,5}=0,2\left(M\right)\)

c) Ta có: \(n_{FeSO_4.7H_2O}=n_{FeSO_4}=0,1\left(mol\right)\)

\(\Rightarrow m_{FeSO_4.7H_2O}=0,1\times278=27,8\left(g\right)\)

d) 4H₂ + Fe₃O₄ \(\underrightarrow{to}\) 3Fe + 4H₂O (2)

Theo Pt2: \(n_{Fe_3O_4}=\frac{1}{4}n_{H_2}=\frac{1}{4}\times0,1=0,025\left(mol\right)\)

\(\Rightarrow m_{Fe_3O_4}=0,025\times232=5,8\left(g\right)\)

\(1.m_{CaCO_3}=400.85\%=340\left(g\right)\\ \rightarrow n_{CaCO_3}=\frac{340}{100}=3,4\left(mol\right)\\ PTHH:CaCO_3+2HCl\rightarrow CaCl_2+H_2O+CO_2\uparrow\\ V_{CO_2}=3,4.22,4=76,16\left(l\right)\\ m_{HCl}=3,4.2.36,5=248,2\left(g\right)\\ m_{CaCl_2}=3,4.111=377,4\left(g\right)\)

\(2.\\ PTHH:C+O_2\underrightarrow{t^o}CO_2\\ PTHH:S+O_2\underrightarrow{t^o}SO_2\\ n_{SO_2}=\frac{4,48}{22,4}=0,2\left(mol\right)\\ \rightarrow m_S=0,2.32=6,4\left(g\right)\\ m_C=\sum_m-m_S=10-6,4=3,6\left(g\right)\\ \%_C=\frac{3,6}{10}.100=36\left(\%\right)\\ n_C=\frac{3,6}{12}=0,3\left(mol\right)\\ \sum n_{O_2}=0,3+0,2=0,5\left(mol\right)\\ \rightarrow V_{O_2}=0,5.22,4=11,2\left(l\right)\\ \rightarrow V_{KK}=5.11,2=56\left(l\right)\\ V_{CO_2}=0,3.22,4=6,72\left(l\right)\)

\(3.\\ n_{CaCO_3}=\frac{200}{100}=2\left(mol\right)\\ PTHH:C+O_2\underrightarrow{t^o}CO_2\\ PTHH:CO_2+Ca\left(OH\right)_2\rightarrow CaCO_3\downarrow+H_2O\\ Theo.pt:n_{CaCO_3}=n_{CO_2}=n_C=2\left(mol\right)\\ \rightarrow m_C=2.12=24\left(g\right)\\ \%_C=\frac{24}{80}.100\%=30\left(\%\right)\\ V_{CO_2}=22,4.2=44,8\left(l\right)\)

\(4.\\ 0,5m^3=500l\\ V_{CH_4}=500-500.2\%=490\left(l\right)\\ PTHH:CH_4+2O_2\underrightarrow{t^o}CO_2+2H_2O\\ n_{CH_4}=\frac{490}{22,4}=21,875\left(mol\right)\\ V_{O_2}=22,4.\left(21,875.2\right)=980\left(l\right)=0,98\left(m^3\right)\)

\(n_{H_2}=\frac{0,672}{22,4}=0,03\left(mol\right)\)

PTHH: \(Ba+2H_2O\rightarrow Ba\left(OH\right)_2+H_2\)

_______a--------------------->a---------->a______(mol)

\(2Na+2H_2O\rightarrow2NaOH+H_2\)

_a----------------------->a------>0,5a_______(mol)

=> a + 0,5a = 0,03

=> a = 0,02 (mol)

PTHH: \(Ba\left(OH\right)_2+2HCl\rightarrow BaCl_2+2H_2O\)

________0,02-------->0,04_______________________(mol)

\(NaOH+HCl\rightarrow NaCl+H_2O\)

_0,02----->0,02_______________________(mol)

=> \(n_{HCl}=0,04+0,02=0,06\left(mol\right)\)

=> C_M(HCl) = \(\frac{0,06}{0,1}=0,6\left(l\right)\)

a) \(n_{Al}=\dfrac{10,8}{27}=0,4\left(mol\right)\)

PTHH: 2Al + 6HCl --> 2AlCl₃ + 3H₂

             0,4------------>0,4---->0,6

=> \(V_{H_2}=0,6.22,4=13,44\left(l\right)\)

b) 

\(m_{AlCl_3}=0,4.133,5=53,4\left(g\right)\)

c)

PTHH: CuO + H₂ --t^o--> Cu + H₂O

                       0,6------>0,6

=> m_Cu = 0,6.64 = 38,4 (g)

`n_[Al]=[10,8]/27=0,4(mol)`

`2Al + 6HCl -> 2AlCl_2 + 3H_2 \uparrow`

`0,4`                              `0,4`     `0,6`      `(mol)`

`a)V_[H_2]=0,6.22,4=13,44(l)`

`b)m_[AlCl_2]=0,4.98=39,2(g)`

`c)`

`H_2 + CuO` $\xrightarrow{t^o}$ `Cu + H_2 O`

`0,4`                     `0,4`                 `(mol)`

`=>m_[Cu]=0,4.64=25,6(g)`