Bài 4 : 

- Cân lấy 32 gam NaCl vào cốc 

- Đong lấy 100 gam nước cho vào cốc, khuấy đều.

Bài 6 : 

\(C\%_{NaCl} = \dfrac{S}{S + 100}.100\% = \dfrac{36}{136}.100\% = 26,47\%\)

Bài 5 : 

Bài 6: ta có mdd=36+100=136( ct-36g, dm là nước 100g)

\(C\%=\dfrac{36}{136}\cdot100\%=26\%\)

a/ \(n_{KOH}=0,2.1=0,2\left(mol\right);n_{H_2SO_4}=0,3.1=0,3\left(mol\right)\)

PTHH: 2KOH + H₂SO₄ → K₂SO₄ + 2H₂O

Mol:        0,2         0,1            0,1

Ta có: \(\dfrac{0,2}{2}< \dfrac{0,3}{1}\) ⇒ KOH hết, H₂SO₄ dư

b/ \(m_{H_2SO_4dư}=\left(0,3-0,1\right).98=19,6\left(g\right)\)

c/ V_{dd sau pứ} = 0,2 + 0,3 = 0,5 (l)

d/ \(C_{M_{ddK_2SO_4}}=\dfrac{0,1}{0,5}=0,2M\)

 \(C_{M_{ddH_2SO_4dư}}=\dfrac{0,3-0,1}{0,5}=0,4M\)

a, \(Zn+2HCl\rightarrow ZnCl_2+H_2\)

b, \(n_{Zn}=\dfrac{6,5}{65}=0,1\left(mol\right)\)

\(m_{HCl}=100.14,6\%=14,6\left(g\right)\Rightarrow n_{HCl}=\dfrac{14,6}{36,5}=0,4\left(mol\right)\)

Xét tỉ lệ: \(\dfrac{0,1}{1}< \dfrac{0,4}{2}\), ta được HCl dư.

Theo PT: \(n_{H_2}=n_{Zn}=0,1\left(mol\right)\Rightarrow V_{H_2}=0,1.22,4=2,24\left(l\right)\)

b, Theo PT: \(\left\{{}\begin{matrix}n_{HCl\left(pư\right)}=2n_{Zn}=0,2\left(mol\right)\\n_{ZnCl_2}=n_{Zn}=0,1\left(mol\right)\end{matrix}\right.\)

\(\Rightarrow n_{HCl\left(dư\right)}=0,4-0,2=0,2\left(mol\right)\)

Ta có: m _{dd sau pư} = 6,5 + 100 - 0,1.2 = 106,3 (g)

\(\Rightarrow\left\{{}\begin{matrix}C\%_{ZnCl_2}=\dfrac{0,1.136}{106,3}.100\%\approx12,79\%\\C\%_{HCl\left(dư\right)}=\dfrac{0,2.36,5}{106,3}.100\%\approx6,87\%\end{matrix}\right.\)

\(n_{MnO_2} = \dfrac{4,35}{87} = 0,05(mol)\)

MnO₂ + 4HCl → MnCl₂ + Cl₂ + 2H₂O

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

\(n_{NaOH} = 0,3.0,1 = 0,03(mol)\)

2NaOH + Cl₂ → NaCl +   NaClO + H₂O

0,03........0,05.....0,015.......0,015....................(mol)

Vậy :

\(C_{M_{NaCl}} = C_{M_{NaClO}} = \dfrac{0,015}{0,3} = 0,05M\)

a) $2NaOH + Cl_2 \to NaCl + NaCl + H_2O$

b) $n_{Cl_2} = \dfrac{6,5}{22,4} = 0,29(mol)$

$n_{NaOH} = 2n_{Cl_2} = 0,58(mol)$

$V_{dd\ NaOH} = \dfrac{0,58}{1} = 0,58(lít)$

$n_{NaCl} = n_{NaClO} = n_{Cl_2} = 0,29(mol)$
$C_{M_{NaCl}} = C_{M_{NaClO}} = \dfrac{0,29}{0,58} = 0,5M$

\(a)n_{NaOH}=0,5.0,2=0,1mol\\ n_{H_2SO_4}=0,3.1=0,3mol\\2 NaOH+H_2SO_4\rightarrow Na_2SO_4+2H_2O\\ \Rightarrow\dfrac{0,1}{2}< \dfrac{0,3}{1}\Rightarrow H_2SO_4.dư\\ 2NaOH+H_2SO_4\rightarrow Na_2SO_4+2H_2O\)

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

\(C_M\) \(_{Na_2SO_4}=\dfrac{0,05}{0,2+0,3}=0,1M\)

\(C_M\) \(_{H_2SO_4}=\dfrac{0,3-0,05}{0,2+0,3}=0,5M\)

b) Vì H₂SO₄ dư nên quỳ tím hoá đỏ.

Vào 20ml dung dịch gì á bạn 

NaOH ấy bạn