a, PT \(H_2SO_4+2NaOH\rightarrow Na_2SO_4+2H_2O\)
b, Ta có: \(m_{H_2SO_4}=\frac{20.24,5}{100}=4,9\left(g\right)\Rightarrow n_{H_2SO_4}=\frac{4,9}{98}=0,05\left(mol\right)\)
Theo PT: \(n_{NaOH}=2n_{H_2SO_4}=0,1\left(mol\right)\)
\(\Rightarrow V_{NaOH}=\frac{0,1}{1}=0,1\left(l\right)\)
c, Ta có: \(m_{ddNaOH}=1,05.0,1=0,105\left(g\right)\)
⇒ m dd sau pư = m dd H2SO4 + m dd NaOH = 20 + 0,105 = 20,105 (g)
Theo PT: \(n_{Na_2SO_4}=n_{H_2SO_4}=0,05\left(mol\right)\)
\(\Rightarrow C\%_{Na_2SO_4}=\frac{0,05.142}{20,105}.100\%\approx35,31\%\)
Bạn tham khảo nhé!