\(a,m_{H_2SO_4}=\dfrac{500.49}{100}=245g\\ n_{H_2SO_4}=\dfrac{245}{98}=2,5\left(mol\right)\\ b,m_{NaOH}=\dfrac{60.20}{100}=12g\\ n_{NaOH}=\dfrac{12}{40}=0,3\left(mol\right)\\ b,m_{CuSO_4}=\dfrac{50.10}{100}=5g\\ n_{CuSO_4}=\dfrac{5}{160}=\dfrac{1}{32}\left(mol\right)\\ d,m_{BaCl_2}=\dfrac{200.5}{100}=10g\\ n_{BaCl_2}=\dfrac{10}{208}=\dfrac{5}{104}\left(mol\right)\)

\(a,n_{H_2}=\dfrac{3,36}{22,4}=0,15\left(mol\right)\)

PTHH: 

Mg + H₂SO₄ ---> MgSO₄ + H₂

         0,15<-------------------0,15

b, m_Mg = 0,15.24 = 3,6 (g)

\(c,n_{Al}=\dfrac{3,6}{27}=\dfrac{2}{15}\left(mol\right)\)

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

           \(\dfrac{2}{15}\)------------------------------------>0,2

So sánh: 0,2 > 0,15 => Al cho nhiều H₂ hơn

pthh  MG + H2SO4 ->MGSO4+ H2

số mol chất khí là n= v/22,4 =0,15 mol

B)khối lượng mgso4 là  m = n .M=0,15.120 = 18g

D

muối tan đc trong nước : NaNO₃ , KCl , K₂CO₃ , CaCl₂ , Na₂SO₄ , 

=> chọn D

28 : D
29 D 
30 C
31 B
32 D 
33 \(S=\dfrac{m_{ct}}{m_{H_2O}}.100\)

D

D. Cả A, B đúng.

Bài 1:

a) Fe + H₂SO₄ --> FeSO₄ + H₂

b) \(n_{Fe}=\dfrac{16,8}{56}=0,3\left(mol\right)\)

PTHH: Fe + H₂SO₄ --> FeSO₄ + H₂

           0,3--->0,3-------->0,3--->0,3

=> V_H2 = 0,3.22,4 = 6,72 (l)

c) m_H2SO4 = 0,3.98 = 29,4 (g)

=> \(C\%_{dd.H_2SO_4}=\dfrac{29,4}{600}.100\%=4,9\%\)

d) m_{dd sau pư} = 16,8 + 600 - 0,3.2 = 616,2 (g)

\(C\%_{FeSO_4}=\dfrac{0,3.152}{616,2}.100\%=7,4\%\)

Bài 2:

a) Zn + 2HCl --> ZnCl₂ + H₂

b) \(n_{Zn}=\dfrac{13}{65}=0,2\left(mol\right)\)

PTHH: Zn + 2HCl --> ZnCl₂ + H₂

            0,2-->0,4----->0,2--->0,2

=> V_H2 = 0,2.22,4 = 4,48 (l)

c) m_HCl = 0,4.36,5 = 14,6 (g)

=> \(C\%_{dd.HCl}=\dfrac{14,6}{400}.100\%=3,65\%\)

d) m_{dd sau pư} = 13 + 400 - 0,2.2 = 412,6 (g)

=> \(C\%_{ZnCl_2}=\dfrac{0,2.136}{412,6}.100\%=6,6\%\)

\(m_{Fe}=\dfrac{11,2}{56}=0,2\left(mol\right)\\ m_{H_2SO_4}=200.12,25\%=24,5g\\ n_{H_2SO_4}=\dfrac{24,5}{98}=0,25\\ pthh:Fe+H_2SO_4\rightarrow FeSO_4+H_2\uparrow\\ LTL:0,2< 0,25=>H_2SO_4\left(d\text{ư}\right)Fe\left(h\text{ết}\right)\) 
    \(n_{H_2}=n_{Fe}=0,2\left(mol\right)\\ V_{H_2}=0,2.22,4=4,48l\)
\(C\%=\dfrac{11,2}{24,5}.100\%=45,7\%\)

`11,2 g` j bạn ?

\(n_{Al}=\dfrac{18,9}{27}=0,7\left(mol\right)\\ pthh:4Al+3O_2\underrightarrow{t^o}2Al_2O_3\) 
           0,7      0,525   0,35 
\(V_{O_2}=\left(0,525.24\right)=12,6l\)
 
\(2Al+3S\underrightarrow{t^o}Al_2S_3\) 
0,7              0,35 
\(m_{Al_2S_3}=0,35.150=52,5g\)

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

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

           0,4-->0,6---------->0,2------->0,6

=> \(C_{M\left(dd.H_2SO_4\right)}=\dfrac{0,6}{0,15}=4M\)

b) V_H2 = 0,6.22,4 = 13,44 (l)

c) \(C_{M\left(Al_2\left(SO_4\right)_3\right)}=\dfrac{0,2}{0,15}=\dfrac{4}{3}M\)

\(n_{Al}=\dfrac{10,8}{27}=0,4\left(mol\right)\\ pthh:2Al+3H_2SO_4\rightarrow Al_2\left(SO_4\right)_3+3H_2\) 
           0,4     0,6                   0,2             0,6 
\(C_M_{H_2SO_4}=\dfrac{0,6}{0,15}=4M\\ V_{H_2}=0,622,4=13,44L\) 
\(C_M=\dfrac{0,2}{0,15}=1,3M\)

\(n_{H_2}=\dfrac{6,72}{22,4}=0,3mol\)

\(Fe+2HCl\rightarrow FeCl_2+H_2\)

0,3       0,6                      0,3     ( mol )

\(m_{Fe\left(pứ\right)}=0,3.56=16,8g\)

\(m_{Fe\left(dư\right)}=28-16,8=11,2g\)

\(m_{HCl}=0,6.36,5=21,9g\)