\(n_{NaCl}=\dfrac{m}{M}=\dfrac{2,925}{\left(23+35,5\right)}=0,05\left(mol\right)\\ PTHH:NaOH+HCl\rightarrow NaCl+H_2O\)
1 1 1 1
0,05 0,05 0,05 0,05
\(m_{NaOH}=n.M=0,05.\left(23+16+1\right)=2\left(g\right)\\ m_{HCl}=n.M=0,05.\left(1+35,5\right)=1,825\\ m_{H_2O}=n.M=0,05.\left(2+16\right)=0,9\left(g\right).\)