\(n_{Fe}=\dfrac{14}{56}=0,25\left(mol\right)\\ pthh:Fe+H_2SO_4\rightarrow FeSO_4+H_2\) 
         0,25                    0,25          0,25 
=> \(V_{H_2}=0,25.22,4=5,6\left(l\right)\) 
   \(m_{FeSO_4}=152.0,25=38\left(g\right)\)  
\(pthh:CuO+H_2\underrightarrow{t^o}H_2O+Cu\)
           0,25     0,25             0,25
=> \(m_{Cu}=0,25.64=16\left(g\right)\)

nFe = 14/56 =0,25 mol

PTHH : Fe + H2SO4 => FeSO4 + H2 (1)

Theo pt(1) : nH2 = nFe = 0,25 mol

VO2 = 0,25 x 22,4 = 5,6 l

Theo pt(1): nFeSO4 = nFe = 0,25 mol

mFeSO4= 0,25 x 152 = 38 g

PTHH : H2 + CuO => Cu +  H2O(2)

theo pt (2) => nH2 = nCu = 0,25 mol

mCu = 0,25 x 64 = 16 g

\(n_{Mg}=\dfrac{4,8}{24}=0,2\left(mol\right)\)

PTHH: Mg + H₂SO4 → MgSO₄ + H₂

Mol:      0,2                                    0,2

\(V_{H_2}=0,2.24,79=4,958\left(l\right)\)

 ⇒ Chọn C

b) m_H2SO4 ban đầu là bao nhiêu :) ?

câu 5

Mg+H2SO4->MgSO4+H2

0,35----------------0,35---0,35

n Mg=0,35 mol

=>VH2=0,35.22,4=7,84l

=>m MgSO4=0,35.120=43g

b)H2+CuO-to>H2O+Cu

0,35----0,35--------------0,35

=>m CuO=0,35.80=28g

=>m Cu=0,35.64=22,4g

Câu 5

`a)n_{Mg} = {8,4}/{24} = 0,35(mol)`

`PTHH : Mg + H_2SO_4 -> MgSO_4 + H_2`

`n_{H_2} = n_{Mg} = 0,35(mol)`

`-> V_{H_2} = 0,35.22,4 = 7,84(gam)`

`n_{MgSO_4} = n_{Mg} = 0,35(mol)`

`-> m_{MgSO_4} = 0,35.120 = 42(gam)`

`b)H_2 + CuO ->^{t^o} Cu + H_2O`

`n_{CuO} = n_{H_2} = 0,35(mol)`

`-> m_{CuO} = 0,35.80 = 28(gam)`

`n_{Cu} = n_{H_2} = 0,35(mol)`

`-> m_{Cu} = 0,35.64 = 22,4(gam)`

Ta có: \(n_{Mg}=\dfrac{2,4}{24}=0,1\left(mol\right)\)

PT: \(Mg+2HCl\rightarrow MgCl_2+H_2\)

_____0,1_____0,2___________0,1 (mol)

a, \(V_{H_2}=0,1.24,79=2,479\left(l\right)\)

b, \(C_{M_{HCl}}=\dfrac{0,2}{1}=0,2\left(M\right)\)

a: \(2Al+6HCl\rightarrow2AlCl_3+3H_2\uparrow\)

0,2               0,6          0,2    0,3

1mol Al sẽ thu được 0,3mol H2

=>0,1 mol Al sẽ thu được 0,03 mol H2

Thể tích khí hidro thu được là:

24,79*0,03=0,7437(lít)

b: m AlCl3=0,2(27+35,5*3)=26,7(g)

\(n_{Al}=\dfrac{5,4}{27}=0,2\left(mol\right)\)

Pt : \(2Al+3H_2SO_4\rightarrow Al_2\left(SO_4\right)_3+3H_2\uparrow\)

       0,2                              0,1         0,3

a) \(V_{H2}=0,3.24,79=7,437\left(l\right)\)

b) \(m_{Al2\left(SO4\right)3}=0,1.342=34,2\left(g\right)\)

 Chúc bạn học tốt

\(a,n_{Al}=\dfrac{10,8}{27}=0,4\left(mol\right)\\ PTHH:2Al+3H_2SO_{4\left(loãng\right)}\rightarrow Al_2\left(SO_4\right)_3+3H_2\uparrow\\ Theo.pt:n_{H_2}=\dfrac{3}{2}n_{Al}=\dfrac{3}{2}.0,4=0,6\left(mol\right)\\ b,PTHH:RO+H_2\underrightarrow{t^o}R+H_2O\\ Mol:0,6\leftarrow0,6\rightarrow0,6\\ M_R=\dfrac{38,4}{0,6}=64\left(\dfrac{g}{mol}\right)\\ \Rightarrow R.là.Cu\)

2Al+3H2SO4->Al2(SO4)3+3H2

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

n Al=0,4 mol

=>VH2=0,6.22,4=13,44l

b)

H2+XO-to>X+H2O

0,6------------0,6

=>0,6=\(\dfrac{38,4}{X}\)

=>X=64 đvC

=>X là Cu(đồng)

=>X=48

\(4.a/n_{Al}=\dfrac{5,4}{27}=0,2mol\\ 2Al+3H_2SO_4\rightarrow Al_2\left(SO_4\right)_3+3H_2\)

0,2       0,3                  0,1                  0,3

\(V_{H_2}=0,3.24,79=7,437l\\ b/C_{\%H_2SO_4}=\dfrac{0,3.98}{150}\cdot100=19,6\%\\ c/m_{Al_2\left(SO_4\right)_3}=0,1.342=34,2g\)

\(5.a/n_{MgO}=\dfrac{4}{40}=0,1mol\\ MgO+2HCl\rightarrow MgCl_2+H_2O\)

0,1            0,2             0,1              0,1

\(C_{\%HCl}=\dfrac{0,2.36,5}{200}\cdot100=3,65\%\\ b/C_{\%MgCl_2}=\dfrac{0,1.95}{200+4}\cdot100=4,66\%\\ c/NaOH+HCl\rightarrow NaCl+H_2O\\ n_{NaOH}=n_{HCl}=0,2mol\\ V_{NaOH}=\dfrac{0,2}{1}=0,2l=200ml\)

\(n_{Al}=\dfrac{2,7}{27}=0,1mol\\ a.2Al+3H_2SO_4\rightarrow Al_2\left(SO_4\right)_3+3H_2\)

0,1           0,15               0,05               0,15

\(b.V_{H_2}=0,15.24,79=3,7185l\\ c.m_{Al_2\left(SO_4\right)_3}=0,05.342=17,1g\\ d.C_{M_{H_2SO_4}}=\dfrac{0,15}{0,4}=0,375M\)

D