a, \(n_{H_2}=\dfrac{2,24}{22,4}=0,1\left(mol\right)\)

PTHH: M + 2H₂O → M(OH)₂ + H₂

Mol:    0,1                     0,1        0,1

\(\Rightarrow M_M=\dfrac{4}{0,1}=40\left(g/mol\right)\)

  ⇒ M là canxi (Ca)

\(C\%_{ddCa\left(OH\right)_2}=\dfrac{0,1.74.100\%}{500}=1,48\%\)

b) \(m_{Ca\left(OH\right)_2}=200.1,48=2,96\left(g\right)\Rightarrow n_{Ca\left(OH\right)_2}=\dfrac{2,96}{74}=0,04\left(mol\right)\)

PTHH: Ca(OH)₂ + 2HCl → CaCl₂ + 2H₂O

Mol:        0,04         0,08

\(V_{ddHCl}=\dfrac{0,08}{2}=0,04\left(l\right)=40\left(ml\right)\)

\(m_{NaOH\left(35\%\right)}=100.35\%=35\left(g\right)\)

\(m_{ddNaOH\left(20\%\right)}=\dfrac{35}{20}.100=175\left(g\right)\)

⇒ m_{nước thêm vào} = 175-100 = 75(g)

_{Vnước thêm vào} = 75.1 = 75 (ml)

Câu 4:

\(n_{HCl}=4.2,75=11\left(mol\right)\)

Ta có: \(V_{ddHCl\left(2M\right)}=\dfrac{n_{HCl\left(1\right)}}{2}\left(l\right);V_{ddHCl\left(3M\right)}=\dfrac{n_{HCl\left(2\right)}}{3}\left(l\right)\)

  \(\Rightarrow V_{ddHCl\left(2M\right)}+V_{ddHCl\left(3M\right)}=\dfrac{n_{HCl\left(1\right)}}{2}+\dfrac{n_{HCl\left(2\right)}}{3}\)

  \(\Leftrightarrow4=\dfrac{3n_{HCl\left(1\right)}+2n_{HCl\left(2\right)}}{6}\Leftrightarrow3n_{HCl\left(1\right)}+2n_{HCl\left(2\right)}=24\)

Ta có: \(\left\{{}\begin{matrix}n_{HCl\left(1\right)}+n_{HCl\left(2\right)}=11\\3n_{HCl\left(1\right)}+2n_{HCl\left(2\right)}=24\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}n_{HCl\left(1\right)}=2\\n_{HCl\left(2\right)}=9\end{matrix}\right.\)

  \(\Rightarrow V_{ddHCl\left(2M\right)}=\dfrac{2}{2}=1\left(l\right);V_{ddHCl\left(3M\right)}=\dfrac{9}{3}=3\left(l\right)\)

Câu 3:

\(m_{NaOH}=35.100\%=35\left(g\right)\\ m_{H_2O\left(thêm\right)}=a\left(g\right)\\ \Rightarrow\dfrac{35}{100+a}.100\%=20\%\\ \Leftrightarrow a=75\left(g\right)\\ \Rightarrow V_{H_2O\left(thêm\right)}=\dfrac{75}{1}=75\left(ml\right)\)

Câu 4:

\(Đặt:V_{ddHCl\left(pha.thêm2M\right)}=a\left(l\right)\\ V_{ddHCl\left(pha.thêm3M\right)}=b\left(l\right)\\ \Rightarrow\dfrac{2a+3b}{a+b}=2,75\left(1\right)\\ Mà:a+b=4\left(2\right)\\ Từ\left(1\right),\left(2\right)\Rightarrow\left\{{}\begin{matrix}a=1\left(l\right)\\b=3\left(l\right)\end{matrix}\right.\)

Vậy cần pha 1 lít dung dịch HCl 2M vào 3 lít dung dịch HCl 3M để thu được 4 lít dung dịch HCl 2,75M 

Chúc em học tốt!

Câu 2 : 

$n_{HCl} = 0,2.1 + 0,3.1,5 = 0,65(mol)$
$V_{dd} = 0,2 + 0,3 = 0,5(mol)$
$C_{M_{HCl}} = \dfrac{0,65}{0,5} =1,3M$

Câu 3 : 

Gọi $m_{H_2O\ cần\ thêm} =a (gam)$

Sau khi thêm : 

$m_{NaOH} = 100.35\% = 35(gam)$
$m_{dd} = 100 + a(gam)$

Suy ra:  $\dfrac{35}{100 + a}.100\% = 20\%$

Suy ra: a = 75(gam)

Câu 4 : 

Gọi $V_{dd\ HCl\ 2M} =a (lít) ; V_{dd\ HCl\ 3M} = b(lít)$

Ta có : 

$a + b = 4$
$2a + 3b = 4.2,75$

Suy ra a = 1(lít) ; b = 3(lít)

Đáp án A

n_HCl= n_H+= 0,4.10^-2 (mol), m_HCl= 0,146 gam

Đáp án A

• V lít HCl 1M + 0,05 mol Na[Al(OH)₄] → 0,01 mol ↓Al(OH)₃

Cần ít nhất HCl → HCl thiếu

HCl + Na[Al(OH)₄] → NaCl + Al(OH)₃↓ + H₂O

n_HCl = 0,01 mol → V_HCl = 0,01 : 1 = 0,01 lít = 10 ml

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

Theo PTHH :

\(n_{HCl} = 2n_{Fe} = 2.\dfrac{11,2}{56} = 0,4(mol)\)

\(NaOH + HCl \to NaCl + H_2O\)

Ta có :

\(n_{NaOH} = n_{HCl} = 0,4(mol)\\ \Rightarrow m_{dd\ NaOH} = \dfrac{0,4.40}{20\%} = 80(gam)\\ \Rightarrow V_{dd\ NaOH} = \dfrac{80}{1,12} = 71,43(ml)\)

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