`#Ly`

Tóm Tắt:

`C_MKOH` `=` `2` $(M)$

`V_(HCl` `=` `250` `ml`

`C_MHCl` `=` `1,5` $(M)$

$a)$ `V_(KOH` `=` $?$

$b)$ `C%_(dd(spư)` `=` $?$

$c)$ Thay dd $KOH$ bằng dd $NaOH$ 

`C%_(NaOH` `=` `10%`

`m_(ddNaOH`cần dùng để trung hòa $HCl$ $=$ $?$

Bài Giải:

$a)$

Đổi $250$ $ml$ $=$ $0,25$ $l$

`n_(HCl` `=` `C_M.V=1,5.0,25=0,375` $(mol)$

PTHH: $KOH$ $+$ $HCl$ `->` $KCl$ $+$ $H_2O$

Theo PT, `n_(KOH` `=` `n_(HCl` `=` `0,375` $(mol)$

`V_(KOH` `=` `n/(C_M)` `=` `(0,375)/(2)=0,1875` $(l)$

$b)$

Theo PT, `n_(KCl` `=` `n_(KOH` `=` `0,375` $(mol)$

`V_(ddKCl` `=` `0,25+0,1875=0,4375` $(l)$

`C_MKCl` `=` `n/V` `=` `(0,375)/(0,4375)≈0,86` $(M)$

$c)$

PTHH: $NaOH$ $+$ $HCl$ `->` $NaCl$ $+$ $H_2O$

Theo PT, `n_(NaOH` `=` `n_(HCl` `=` `0,375` $(mol)$

`m_(NaOH` `=` $n.M=0,375.40=15$ $(g)$

`m_(ddNaOH` `=` `(m_{NaOH}.100)/(C%_{NaOH})` `=` `(15.100)/(10)` `=` `150` $(g)$

Đáp án + Giải thích các bước giải:

Tóm tắt:
$C_{MKOH}= 2M$

$V_{HCl} = 250ml = 0,25l$

$C_{MHCl} = 1,5M$

a) $V_{KOH} = ?l$

b) $C_{Mdd\text{muối}} = ?M$

c) Thay $KOH$ bằng $NaOH$

$C\%_{NaOH} = 10\%$

$m_{ddNaOH} = ?(g)$

Giải:

 a)

$n_{HCl} = C_M . V  = 1,5 . 0,25 = 0,375(mol)$
PTHH:
$KOH + HCl \longrightarrow KCl + H_2O$

Theo PTHH:
$n_{KOH} = n_{HCl} = 0,375(mol)$
$\rightarrow V_{KOH} = \dfrac{n}{C_M} = \dfrac{0,375}{2} = 0,1875(l) = 187,5(ml)$
b) $V_{KCl} = 0,1875 + 0,25 = 0,4375(l)$

Theo PTHH:
$n_{KCl} = n_{HCl} = 0,375(mol)$
$\rightarrow C_{MKCl} = \dfrac{n}{V} = \dfrac{0,375}{0,4375} = \dfrac{6}{7}M \approx 0,86M$

c) PTHH:
$NaOH + HCl \longrightarrow NaCl + H_2O$

Theo PTHH:
$n_{NaOH} = n_{HCl} = 0,375(mol)$
$\rightarrow m_{NaOH} = n . M = 0,375 . (23 + 16 + 1) = 0,375 . 40 = 15(g)$

$\rightarrow m_{ddNaOH} = \dfrac{m_{ct} . 100\%}{C\%} = \dfrac{15 . 100\%}{10\%} =150(g)$