nNaOH=4:40=0,1(mol)
Ta có PTHH:
NaOH+HCl->NaCl+H2O
0,1.........0,1......0,1..............(mol)
Ta có:\(\dfrac{n_{NaOH}}{1}< \dfrac{n_{HCl}}{1}\)(0,1<0,15)=>NaOH hết,HCl dư.Tính theo NaOH
Theo PTHH:
Sau pư,chất tan gồm:
mHCl(dư)=(0,15-0,1).36,5=1,825(g)
mNaCl=0,1.58,5=5,85(g)
=>m=mct=mHCl(dư)+mNaCl=5,85+1,825=7,675(g)