Question

Cho 100g dung dịch NaOH 4% tác dụng vừa đủ dung dịch HCl 3,65%

a) mHCl phản ứng 

b) mdd HCl

c) mNaCl

Answer 1

\(m_{NaOH}=100.4\%=4\left(g\right)\)

\(n_{NaOH}=4:40=0,1\left(mol\right)\)

PTHH:

\(NaOH+HCl\rightarrow NaCl+H_2O\)

0,1            0,1           0,1

\(a,m_{HCl}=0,1.36,5=3,65\left(g\right)\)

\(b,m_{ddHCl}=3,65:3,56\%=100\left(g\right)\)

\(c,m_{NaCl}=0,1.58,5=5,85\left(g\right)\)

Answer 2

mNaOH=4g=> nNaOH=0,1 mol

 

 

NaOH+ HCl-> NaCl+ H2O

 

0,1         0,1          0,1 

 

mHCl=0,1x36,5=3,65g

 

x=3,65x100:3,65=100 g

 

mNaCl=5,85g

 

mdd=100+100=200g

 

C%=5,85:200x100%=2,925%