Bài 2  :

$m_{HCl}=7,3\%.50=3,65g \\⇒n_{HCl}=\dfrac{3,65}{36,5}=0,1mol \\a.PTHH : \\Zn+2HCl\to ZnCl_2+H_2↑ \\b.Theo\ pt : \\n_{Zn}=\dfrac{1}{2}.n_{HCl}=\dfrac{1}{2}.0,1=0,05mol \\⇒m=m_{Zn}=0,05.65=3,25g \\c.Theo\ pt : \\n_{H_2}=\dfrac{1}{2}.n_{HCl}=\dfrac{1}{2}.0,1=0,05mol \\⇒V_{H_2}=0,05.22,4=1,12l \\d.Theo\ pt : \\n_{ZnCl_2}=\dfrac{1}{2}.n_{HCl}=\dfrac{1}{2}.0,1=0,05mol  \\⇒m_{ZnCl_2}=0,05.136=6,8g$

Bài 3 :

$n_{Zn}=\dfrac{6,5}{65}=0,1mol \\a.PTHH : \\Zn+2HCl\to ZnCl_2+H_2↑ \\b.Theo\ pt : \\n_{HCl}=2.n_{Zn}=2.0,1=0,2mol \\⇒V=\dfrac{0,2}{2}=0,1l=100ml \\c.Theo\ pt : \\n_{H_2}=n_{Zn}=0,1mol \\⇒V_{H_2}=0,1.22,4=2,24l \\d.Theo\ pt : \\n_{ZnCl_2}=n_{Zn}=0,1mol \\⇒m_{ZnCl_2}=0,1.136=13,6g$

Bài 1 :

$m_{muối(1)}=50.20\%=10g \\m_{muối(2)}=50.5\%=2,5g \\⇒m_{muối(3)}=10+2,5=12,5g \\m_{dd}=50+50=100g \\⇒C\%_{muối(3)}=\dfrac{12,5}{100}.100\%=12,5\%$

Bài 4 :

$n_{dd\ đường\ 0,5M}=2,5.0,5=1,25mol \\n_{dd\ đường\ 1M}=3.1=3mol \\V_{dd}=2,5+3=5,5l \\⇒C_{M\ dd\ đường}=\dfrac{1,25+3}{5,5}≈0,77mol$

$Bài 5 : m_{HCl}=7,3\%.50=3,65g \\⇒n_{HCl}=\dfrac{3,65}{36,5}=0,1mol \\a.PTHH : \\Zn+2HCl\to ZnCl_2+H_2↑ \\b.Theo\ pt : \\n_{H_2}=\dfrac{1}{2}.n_{HCl}=\dfrac{1}{2}.0,1=0,05mol \\⇒V_{H_2}=0,05.22,4=1,12l \\c.Theo\ pt : \\n_{Zn}=\dfrac{1}{2}.n_{HCl}=\dfrac{1}{2}.0,1=0,05mol \\⇒m=m_{Zn}=0,05.65=3,25g$

Bài 4:

$n_{1}=0,5.2,5=1,25(mol)$

$n_{2}=1.3=3(mol)$

$n_{3}=1,25+3=4,25(mol)$

Thể tích dd sau khi trộn: $V_{3}=2,5+3=5,5(l)$

$C_{M}=$ $\frac{4,25}{5,5}=0,8M$

Bài 5:

$2HCl + Zn → H2 + ZnCl2_{}$

$b)m_{ct_{HCl}}=$ $\frac{50.7,3}{100}=3,65(g)$

$n_{HCl}=$ $\frac{3,65}{36,5}=0,1(mol)$

$2HCl + Zn → H2 + ZnCl2_{}$

   2.............1...........1..........1

  0,1..........0,05.....0,05....0,05 (mol)

Thể tích khí $H_{2}$ ở(đktc) là: $V_{H2}=0,05.22,4=1,12(l)$ 

$c)_{}$ Khối lượng kẽm là: $m_{Zn}=0,05.65=72,8(g)$