\(2NaCl+H_2SO_4\rightarrow Na_2SO_4+2HCl\)
15299_____________________15299
\(m_{dd_{HCl}}=1250.100.1,19=1487500\)
\(m_{HCl}=1487500.37\%=550375\)
\(n_{HCl}=\frac{550375}{36,5}=15079\)
\(n_{NaCl}=\frac{0,895.1000.1000}{58,5}=15299\)
\(\Rightarrow H=\frac{15079}{15299}.100=98,56\%\)