Đáp án:

`C_{M\ dd\ HCl}=0,5M`

`m_{KCl}=11,66(g)`

`%m_{K_2CO_3}=75,27%\ ; %m_{KCl}=24,73%`

Giải thích các bước giải:

`a)`

`Pt: K_2CO_3+2HCl->2KCl+CO_2+H_2O`

`b)`

`1344(ml)=1,344(l)\ ; 240(ml)=0,24(l)`

`n_{CO_2}={1,344}/{22,4}=0,06(mol)`

`Theo\ pt: n_{HCl}=2n_{CO_2}=0,12(mol)`

`C_{M\ dd\ HCl}={0,12}/{0,24}=0,5M`

`c)`

`Theo\ pt: n_{KCl}=n_{HCl}=0,12(mol)`

`Theo\ pt: n_{K_2CO_3}=n_{CO_2}=0,06(mol)`

`m_{KCl}=(0,12.74,5)+(11-0,06.138)=11,66(g)`

`d)`

`%m_{K_2CO_3}={0,06.138}/{11}.100%=75,27%`

`%m_{KCl}=100%-75,27%=24,73%`

 

$\text{Đáp án + giải thích các bước giải :}$

$Bài$ $16$ $:$

$a/$

$PTHH$ $:$

$K_2CO_3$ $+$ $2HCl$ $→$ $2KCl$ $+$ $CO_2↑$ $+$ $H_2O$ 

$b/$

$n_{CO_2}$ $=$ $\frac{1,344}{22,4}$ $=$ $0,06$ $(mol)$

$Theo$ $pt$ $:$ $n_{HCl}$ $=$ $2n_{CO_2}$ $=$ $0,12$ $(mol)$

$⇒$ $C_{M_{HCl}}$ $=$ $\frac{0,12}{0,24}$ $=$ $0,5$ $M$

$c/$

$Theo$ $pt$ $:$ $n_{KCl}$ $=$ $2n_{CO_2}$ $=$ $0,12$ $(mol)$

$Theo$ $pt$ $:$ $n_{K_2CO_3}$ $=$ $n_{CO_2}$ $=$ $0,06$ $(mol)$

$⇒$ $m_{K_2CO_3}$ $=$ $0,06$ $×$ $138$ $=$ $8,28$ $(g)$

$⇒$ $m_{KCl}$ $=$ $0,12$ $×$ $74,5$ $+$ $(11$ $-$ $8,28)$ $=$ $11,66$ $(g)$

$d/$

$\%_{K_2CO_3}$ $=$ $75,27\%$

$\%_{KCl}$ $=$ $24,73\%$

$\text{Chúc bn hk tốt !! :))))}$