Bài 1:

\(Na_2O+H_2O\rightarrow2NaOH\\ SO_2+H_2O⇌H_2SO_3\\ HCl+KOH\rightarrow KCl+H_2O\\ Na_2O+2HCl\rightarrow2NaCl+H_2O\\ SO_2+2KOH\rightarrow K_2SO_3+H_2O\\ KOH+SO_2\rightarrow KHSO_3\\ CaCO_3+2HCl\rightarrow CaCl_2+CO_2+H_2O\\ SO_2+Na_2O\rightarrow Na_2SO_3\)

Bài 2 nếu m_HCl= 7,3(g) thì đúng hơn 7,1(g) nên anh sửa là 7,3 gam nha em.

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Theo ĐLBTKL,ta có:

\(m_{Zn}+m_{HCl}=m_{ZnCl_2}+m_{H_2}\\ \Leftrightarrow6,5+7,3=m_{ZnCl_2}+0,2\\ \Leftrightarrow m_{ZnCl_2}=\left(6,5+7,3\right)-0,2=13,6\left(g\right)\)

a) $n_{Mg} = \dfrac{4,8}{24} = 0,2(mol)$
$Mg + 2HCl \to MgCl_2 + H_2$

Theo PTHH : 

$n_{HCl} = 2n_{Mg} = 0,4(mol) \Rightarrow m_{HCl} = 0,4.36,5 = 14,6(gam)$

b)

$n_{MgCl_2} = n_{Mg} = 0,2(mol) \Rightarrow m_{MgCl_2} = 0,2.95 = 19(gam)$

\(n_{Mg}=\dfrac{4,8}{24}=0,2\left(mol\right)\\ Mg+2HCl\rightarrow MgCl_2+H_2\\ 0,2.......0,4........0,2.........0,2\left(mol\right)\\ a.m_{HCl}=0,4.36,5=14,6\left(g\right)\\ b.m_{MgCl_2}=0,2.95=19\left(g\right)\)

PT: \(Fe+2HCl\rightarrow FeCl_2+H_2\)

Ta có: \(n_{Fe}=\dfrac{11,2}{56}=0,2\left(mol\right)\)

Theo PT: \(\left\{{}\begin{matrix}n_{H_2}=n_{FeCl_2}=n_{Fe}=0,2\left(mol\right)\\n_{HCl}=2n_{Fe}=0,4\left(mol\right)\end{matrix}\right.\)

a, Ta có: \(V_{H_2}=0,2.22,4=4,48\left(l\right)\)

b, \(V_{ddHCl}=\dfrac{0,4}{1,5}\approx0,267\left(l\right)\)

c, \(m_{FeCl_2}=0,2.127=25,4\left(g\right)\)

Bạn tham khảo nhé!

\(2Al+6HCl->2AlCl_3+3H_2\\ Fe+2HCl->FeCl_2+H_2\\ n_{Al}=a;n_{Fe}=b\\ 27a+56b=8,3\\ 1,5a+b=\dfrac{5,6}{22,4}=0,25\\ a=b=0,1\\ m_{Al}=27\cdot0,1=2,7g\\ m_{Fe}=8,3-2,7=5,6g\\ a=\dfrac{3a+2b}{500}\cdot36,5=3,65\%\\ m_{ddsau}=508,3-0,25\cdot2=507,8g\\ C\%_{AlCl_3}=\dfrac{133,5a}{507,8}=2,63\%\\ C\%_{FeCl_2}=\dfrac{127b}{507,8}=2,50\%\)

a)

Gọi $n_{Zn} = a(mol) ; n_{Al} = b(mol) \Rightarrow 65a + 27b = 11,9(1)$

$Zn + 2HCl \to ZnCl_2 + H_2$
$2Al + 6HCl \to 2AlCl_3 + 3H_2$
Theo PTHH : 

$n_{H_2} = a + 1,5b = \dfrac{8,96}{22,4} = 0,4(2)$

Từ (1)(2) suy ra : a = 0,1; b = 0,2

$m_{Zn} = 0,1.65 = 6,5(gam)$

$m_{Al} = 0,2.27 = 5,4(gam)$

b) $n_{HCl} = 2n_{H_2} = 0,8(mol)$
$C\%_{HCl} = \dfrac{0,8.36,5}{125}.100\% = 23,36\%$

nAl = 5,4 / 27 = 0,2(mol)

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

0,2      0,3                0,1                0,3    (mol)

VH2SO4  = n/ CM = 0,3 / 2 = 0,15(l)

=> V1 = 150 ml

mAl2(SO4)3 = 0,1 . 342 = 34,2 (g)

Al2(SO4)3 + 3BaCl2 -- > 3BaSO4 + 2AlCl3

 0,1                                       0,1

=> mBaSO4 = 0,1 . 233 = 23,3 (g)

Đáp án  B

Đáp án A

Số mol  H 2 thu được là:  n H 2 = 2 , 24 22 , 4 = 0 , 1   mol

Sơ đồ phản ứng hóa học:

Chọn B