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)
\(m_{NaOH}=\dfrac{200.10}{100}=20\left(g\right)\\ n_{NaOH}=\dfrac{20}{40}=0,5\left(mol\right)\)
\(PTHH:HCl+NaOH\rightarrow NaCl+H_2O\)
\(n_{HCl}=n_{NaOH}=0,5\left(mol\right)\\ m_{HCl}=0,5.36,5=18,25\left(g\right)\\ m_{ddHCl}=\dfrac{18,25.100}{3,65}=500\left(g\right)\)
mNaOH=200.10%=20(g)
=>nNaOH\(=\dfrac{20}{40}\)=0,5(mol)
PTHH: NaOH + HCl -> NaCl + H2O
Mol: 0,5 0,5
=> mHCl=0,5.36,5= 18,25(g)
\(m_{ddHCl}=\dfrac{18,25.100}{3,65}=500\left(g\right)\)
