Ta có: \(n_{H_2}=\dfrac{6,72}{22,4}=0,3\left(mol\right)\)

a. PTHH: \(Zn+H_2SO_4--->ZnSO_4+H_2\uparrow\left(1\right)\)

b. Theo PT₍₁₎: \(n_{Zn}=n_{H_2}=0,3\left(mol\right)\)

\(\Rightarrow m_{Zn}=65.0,3=19,5\left(g\right)\)

c. Theo PT₍₁₎: \(n_{H_2SO_4}=n_{Zn}=0,3\left(mol\right)\)

Đổi 300ml = 0,3 lít

\(\Rightarrow C_{M_{H_2SO_4}}=\dfrac{0,3}{0,3}=1M\)

d. PTHH: \(2NaOH+H_2SO_4--->Na_2SO_4+2H_2O\left(2\right)\)

Theo PT₍₂₎: \(n_{NaOH}=2.n_{H_2SO_4}=2.0,3=0,6\left(mol\right)\)

\(\Rightarrow m_{NaOH}=0,6.40=24\left(g\right)\)

\(\Rightarrow m_{dd_{NaOH}}=\dfrac{24.100\%}{20\%}=120\left(g\right)\)

\(n_{Al}=\dfrac{10,8}{27}=0,4(mol)\\ a,2Al+3H_2SO_4\to Al_2(SO_4)_3+3H_2\\ \Rightarrow n_{H_2}=n_{H_2SO_4}=0,6(mol);n_{Al_2(SO_4)_3}=0,2(mol)\\ b,V_{H_2}=0,6.22,4=13,44(l)\\ c,m_{dd_{H_2SO_4}}=\dfrac{0,6.98}{9,8\%}=600(g)\\ \Rightarrow C\%_{Al_2(SO_4)_3}=\dfrac{0,2.342}{10,8+600-0,6.2}.100\%=11,22\%\)

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

Phương trình hóa học phản ứng

2Al + 3H₂SO₄ ---> Al₂(SO₄)₃ + 3H₂

_{2          :   3                   : 1               : 3} 

0,4        0,6            0,2             0,6

mol       mol           mol             mol

Thể tích khí H₂ sinh ra là 

\(V=n.22,4=0,6.22,4=13,44\left(l\right)\)

\(m_{Al_2\left(SO_4\right)_3}=n.M=0,2.342=68,4\left(g\right)\)

\(m_{H_2SO4}=n.M=0,6.98=58,8\left(g\right)\)

\(m_{H_2}=n.M=0,6.2=1,2\left(g\right)\)

=> \(m_{ddH_2SO_4}=\dfrac{m_{H_2SO_4}.100\%}{C\%}=\dfrac{58,8.100\%}{9,8\%}=600\)(g)

=> \(m_{\text{dd sau pư}}=m_{ddH_2SO_4}+m_{Al}-m_{H_2}\)

= 600 + 10,8 - 1,2 (g) = 609,6 (g) 

=> \(C\%_{Al_2\left(SO_4\right)_3}=\dfrac{m_{Al_2\left(SO_4\right)_3}}{m_{\text{dd sau pư}}}.100\%=\dfrac{68,4}{609,6}.100\%\)=11,22%

2Al+3H₂SO₄->Al₂(SO₄)₃+3H₂

            0,1----------\(\dfrac{1}{30}\)-------0,1 mol

n _H2=\(\dfrac{2,24}{22,4}\)=0,1 mol

=>m _Al2(SO4)3=\(\dfrac{1}{30}\).342=11,4g

=>CM _H2SO4=\(\dfrac{0,1}{0,2}\)=0,5 M

giúp em với mn ơi

$Zn+2HCl\rightarrow ZnCl_2+H_2$

$n_{Zn}=\dfrac{6,5}{65}=0,1\ mol$

a) Theo phương trình:

$n_{H_2}=n_{Zn}=0,1\ mol$

$V_{H_2}=0,1\cdot22,4=2,24\ lít$

b) Theo phương trình:

$n_{HCl}=2n_{Zn}=0,2\ mol$

$V_{dd\ HCl}=100\ ml=0,1\ lít$

$C_M(HCl)=\dfrac{0,2}{0,1}=2\ M$

c) Phương trình phản ứng trung hòa:

$Ca(OH)_2+2HCl\rightarrow CaCl_2+2H_2O$

Số mol HCl trong $50\ ml$ dung dịch:

$n_{HCl}=2\cdot0,05=0,1\ mol$

Theo phương trình:

$n_{Ca(OH)_2}=\dfrac{n_{HCl}}{2}=\dfrac{0,1}{2}=0,05\ mol$

Thể tích dung dịch $Ca(OH)_2$:

$V_{Ca(OH)_2}=\dfrac{0,05}{2}=0,025\ lít$

$V=0,025\cdot1000=25\ ml$

1)

\(n_{Al}=\dfrac{2,7}{27}=0,1\left(mol\right)\)

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

____0,1----->0,15

=> m_H2SO4 = 0,15.98 = 14,7(g)

=> \(C\%=\dfrac{14,7}{250}.100\%=5,88\%\)

2)

\(n_{Na_2CO_3}=\dfrac{21,2}{106}=0,2\left(mol\right)\)

PTHH: Na₂CO₃ + 2HCl --> 2NaCl + CO₂ + H₂O

_______0,2------------------------------>0,2

=> V_CO2 = 0,2.22,4 = 4,48(l)

3)

\(n_A=\dfrac{18,4}{M_A}\left(mol\right)\)

PTHH: 2A + Cl₂ --t^o--> 2ACl

____\(\dfrac{18,4}{M_A}\)---------->\(\dfrac{18,4}{M_A}\)

=> \(\dfrac{18,4}{M_A}\left(M_A+35,5\right)=46,8=>M_A=23\left(Na\right)\)

4)

n_HCl = 0,2.3 = 0,6(mol)

PTHH: M + 2HCl --> MCl₂ + H₂

____0,3<-----0,6

=> \(M_M=\dfrac{7,2}{0,3}=24\left(Mg\right)\)

a, PT: \(Zn+H_2SO_4\rightarrow ZnSO_4+H_2\)

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

Theo PT: \(n_{Zn}=n_{H_2SO_4}=n_{ZnCl_2}=n_{H_2}=0,15\left(mol\right)\)

b, m_Zn = 0,15.65 = 9,75 (g)

c, C_{M (H2SO4)} = 0,15/0,05 = 3 M

d, m_ZnSO4 = 0,15.161 = 24,15 (g)

Bạn tham khảo nhé!

\(n_{Mg}=\dfrac{12}{24}=0,5\left(mol\right)\)

PT: \(Mg+H_2SO_4\rightarrow MgSO_4+H_2\)

a, \(n_{H_2}=n_{Mg}=0,5\left(mol\right)\)

\(\Rightarrow V_{H_2}=0,5.22,4=11,2\left(l\right)\)

b, \(n_{H_2SO_4}=n_{Mg}=0,5\left(mol\right)\)\(\Rightarrow C_{M_{H_2SO_4}}=\dfrac{0,5}{0,2}=2,5\left(M\right)\)

a, \(Fe+H_2SO_4\rightarrow FeSO_4+H_2\)

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

Theo PT: \(n_{H_2}=n_{H_2SO_4}=n_{Fe}=0,1\left(mol\right)\)

\(\Rightarrow V_{H_2}=0,1.22,4=2,24\left(l\right)\)

c, \(C_{M_{H_2SO_4}}=\dfrac{0,1}{0,1}=1\left(M\right)\)

a)

$n_{Al} = 0,3(mol)$

$2Al + 3H_2SO_4 \to Al_2(SO_4)_3 + 3H_2$

Theo PTHH : 

$n_{H_2SO_4} = \dfrac{3}{2}n_{Al} = 0,45(mol)$
$m_{dd\ H_2SO_4} = \dfrac{0,45.98}{12,25\%} = 360(gam)$

b)

$n_{H_2} = n_{H_2SO_4} = 0,45(mol)$
$V_{H_2} = 0,45.22,4 = 10,08(lít)$

c)

$n_{Al_2(SO_4)_3} = 0,15(mol)$
$m_{dd\ sau\ pư} = 8,1 + 360 - 0,45.2 = 367,2(gam)$
$C\%_{Al_2(SO_4)_3} = \dfrac{0,15.342}{367,2}.100\% = 14\%$