Question

Cần dùng bao nhiêu gam dd HCL 3,65% để trung hòa 200 gam dd NAOH 10©

Answer 1

 NAOH 10© ?

 

Answer 2

mNaOH=200.10%=20(g)

=>nNaOH=20/40=0,5(mol)

PTHH: NaOH + HCl -> NaCl + H2O

nHCl=nNaOH=0,5(mol) => mHCl=0,5 x 36,5= 18,25(g)

mddHCl=(18,25.100)/3,65= 500(g)

Answer 3

\(m_{NaOH}=\dfrac{200.10}{100}=20\left(g\right)\\ n_{NaOH}=\dfrac{20}{40}=0,5\left(mol\right)\)

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

\(n_{HCl}=n_{NaOH}=0,5\left(mol\right)\\ m_{HCl}=0,5.36,5=18,25\left(g\right)\\ m_{ddHCl}=\dfrac{18,25.100}{3,65}=500\left(g\right)\)

Answer 4

m_NaOH=200.10%=20(g)

=>n_NaOH\(=\dfrac{20}{40}\)=0,5(mol)

PTHH: NaOH + HCl -> NaCl + H₂O

Mol:       0,5        0,5

 => m_HCl=0,5.36,5= 18,25(g)

\(m_{ddHCl}=\dfrac{18,25.100}{3,65}=500\left(g\right)\)