`#AkaShi`
Gửi bạn 😼
.............................
PTHH: `NaOH + HCl -> NaCl + H_2O`
`→n_{NaOH}=(200xx20%)/40=1 (mol)`
`→n_{HCl}=(200xx36,5%)/(36,5)=2 (mol)`
Xét hết dư:
`n_{NaOH}=1 (mol) < n_{HCl}=2 (mol)`
`→HCl` dư ; `NaOH` hết
`→n_{HCl\ pư}=n_{NaOH}=1 (mol)`
`→n_{HCl\ dư}=n_{HCl\ bđ}-n_{HCl\ pư}=2-1=1 (mol)`
`→m_{HCl\ dư}=1xx36,5=36,5 (g)`
`→n_{NaCl}=n_{NaOH}=1 (mol)`
`→m_{NaCl}=1xx58,5=58,5 (g)`
`→m_{dd\ sau\ pư}=m_{NaOH}+m_{HCl}=200+200=400 (g)`
`→C%_{HCl\ dư}=(36,5)/400xx100%=9,125%`
`→C%_{NaCl}=(58,5)/400xx100%=14,625%`