a, Ta có : \(C\%=\frac{m_{ct}}{m_{dd}}.100\%=15\%=\frac{m_{NaOH}}{250}.100\%\)
=> \(m_{NaOH}=37,5\left(g\right)\)
=> \(n_{NaOH}=\frac{m}{M}=\frac{37,5}{40}=0,9375\left(mol\right)\)
PTHH : \(NaOH+HCl\rightarrow NaCl+H_2O\)
.............0,9375........0,9375..............................
=> \(m_{HCl}=n.M=0,9375.36,5=34,21875\left(g\right)\)
Ta có : \(C\%=\frac{m_{ct}}{m_{dd}}.100\%=14,6\%=\frac{34,21875}{m_{dd}}.100\%\)
=> \(m_{ddHCl}=234,375\left(g\right)\)
b, PTHH : \(2NaOH+H_2SO_4\rightarrow Na_2SO_4+2H_2O\)
.................0,9375..........0,46875......................
Ta có : \(C_M=\frac{n}{V}=0,5=\frac{0,46875}{V}\)
=> \(V_{H_2SO_4}=0,9375\left(l\right)\)
