\(NaOH+HCl-->NaCl+H2O\)
\(n_{HCl}=0,2.2=0,4\left(mol\right)\)
\(n_{NaOH}=n_{HCl}=0,4\left(mol\right)\)
\(V_{NaOH}=\frac{0,4}{1}=0,4\left(l\right)\)
b) \(n_{NaCl}=n_{HCl}=0,4\left(mol\right)\)
\(m_{NaCl}=0,4.58,5=23,4\left(g\right)\)
ta có pt NaOH+HCl−−>NaCl+H2O
............0,4--------0,4--------0,4 mol
nHCl=0,2.2=0,4(mol)
---->VNaOH=0,41=0,4(l)
b)
-->mNaCl=0,4.58,5=23,4(g)