\(n_{NaOH}=\dfrac{40}{40}=1\left(mol\right)\)
PTHH:
\(NaOH+HCl\rightarrow NaCl+H_2O\)
1-------->1------>1
\(m_{NaCl}=1.58,5=58,5\left(g\right)\\ m_{HCl}=1.36,5=36,5\left(g\right)\)
NaOH + HCl -> NaCl + H2O
Só mol NaOH \(\dfrac{40}{23+16+1}=1mol\)
=> số mol NaCl = số mol HCl = 1 mol
Khối lượng NaCl : \(\left(23+35,5\right).1=58,5g\)
Khối lượng HCl : \(\left(1+35,5\right).1=36,5g\)