HCl +NAOH----<NaCl + H2O
Ta có
n\(_{HCl}=\frac{10}{36,5}=0,27\left(mol\right)\)
n\(_{NaOH}=\frac{10}{40}=0,25\left(mol\right)\)
=> HCl dư
Theo pthh
n\(_{NaCl}=n_{NaOH}=0,25\left(mol\right)\)
m\(_{NaC_{ }l}=0,25.58,5=14,625\left(g\right)\)
nHCl = 0,27(mol)
nNaOH = 0,25(mol)
PTHH : NaOH+ HCl--->NaCl+H2O
=>nNaCl = 0,25(mol)
=>mNaCl = 0,25.58,5 = 14,62