Đề thiếu rồi em !

\(C_{M_{KOH}}=\dfrac{0.5\cdot1+2a}{0.5+a}=1.2\left(M\right)\)

\(\Leftrightarrow a=0.125\)

Ta có : 

$V_{dd} = 0,5 + a(lít)$

$n_{KOH} = 0,5.1 + a.2 = 2a + 0,5(mol)$
Suy ra: 

$C_{M_{KOH}} = \dfrac{2a + 0,5}{0,5 + a} = 1,2M$
$\Rightarrow a = 0,125(lít)$

a ) \(PTHH:Na_2O+H_2O\rightarrow2NaOH\)

b ) \(n_{Na_2O}=0,25\left(mol\right)\)

\(PT:Na_2O+H_2O\rightarrow2NaOH\)

\(0,25\) \(0,25\) \(\left(mol\right)\)

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

\(0,5\) \(0,5\) \(\left(mol\right)\)

\(V_{HCl}=\frac{0,5}{1}=0,5\left(lít\right)\)

a ) PTHH : \(Na_2O+H_2O\rightarrow2NaOH\)

b ) \(PT:Na_2O+H_2O\rightarrow2NaOH\)

                \(0,25\)                      \(0,5\)    ( mol )

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

                 \(0,5\)          \(0,5\)                   ( mol )

\(V_{HCl}=\frac{0,5}{1}=0,5\left(lít\right)\)

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

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

⇒ 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!

\(V_{dd_{HCl\left(2M\right)}}=a\left(l\right),V_{dd_{HCl\left(3M\right)}}=b\left(l\right)\)

\(\Rightarrow V_{dd}=a+b=4\left(l\right)\left(1\right)\)

\(n_{HCl}=2a+3b=4\cdot2.75=11\left(mol\right)\left(2\right)\)

\(\left(1\right),\left(2\right):a=1,b=3\)