Đáp án:
$C_{M_{HCl}}=0,4M$
$\%_{K_2CO_3}=55,2\%$
$\%_{KCl}=44,8\%$
$m_{KClspứ}=5,96g$
Giải thích các bước giải:
$PTHH$
$K_2CO_3+2HCl→2KCl+CO_2↑+H_2O$
$KCl+HCl→$Phản ứng không xảy ra
$n_{CO_2}=\dfrac{V_{CO_2}}{22,4}=\dfrac{0,896}{22,4}=0,04mol$
Từ phương trình
$→n_{HCl}=0,04.2=0,08mol$
$C_{M_{HCl}}=\dfrac{n_{HCl}}{V_{ddHCl}}=\dfrac{0,08}{0,2}=0,4M$
$n_{K_2CO_3}=n_{CO_2}=0,04mol$
$m_{K_2CO_3}=n_{K_2CO_3}.M_{K_2CO_3}=0,04.138=5,52g$
$m_{KCl}=m_{hh}-m_{K_2CO_3}=10-5,52=4,48g$
$\%_{K_2CO_3}=\dfrac{m_{K_2CO_3}}{m_{hh}}.100\%=\dfrac{5,52}{10}.100\%=55,2\%$
$\%_{KCl}=100\%-55,2\%=44,8\%$
$n_{KClspứ}=0,08mol$
$m_{KClspứ}=n_{KClspứ}.M_{KCl}=0,08.74,5=5,96g$