a, PT: \(NaOH+HCl\rightarrow NaCl+H_2O\)
b, Ta có: \(n_{NaOH}=\dfrac{60}{40}=1,5\left(mol\right)\)
Theo PT: \(n_{NaCl}=n_{NaOH}=1,5\left(mol\right)\)
\(\Rightarrow m_{NaCl}=1,5.58,5=87,75\left(g\right)\)
c, Theo PT: \(n_{HCl}=n_{NaOH}=1,5\left(mol\right)\)
\(\Rightarrow C_{M_{HCl}}=\dfrac{1,5}{0,15}=10\left(M\right)\)