2NaOH + H2SO4 → Na2SO4 + 2H2O
\(n_{NaOH}=\dfrac{4}{40}=0,1\left(mol\right)\)
a) Theo PT: \(n_{H_2SO_4}=\dfrac{1}{2}n_{NaOH}=\dfrac{1}{2}\times0,1=0,05\left(mol\right)\)
\(\Rightarrow C_{M_{H_2SO_4}}=\dfrac{0,05}{0,1}=0,5\left(M\right)\)
b) Theo PT: \(n_{Na_2SO_4}=\dfrac{1}{2}n_{NaOH}=\dfrac{1}{2}\times0,1=0,05\left(mol\right)\)
\(\Rightarrow m_{Na_2SO_4}=0,05\times142=7,1\left(g\right)\)
nNaOH = \(\dfrac{4}{40}=0,1mol\)
2NaOH + H2SO4 -> Na2SO4 +2 H2O
0,1------>0,05------>0,05
a) CMH2SO4 = \(\dfrac{0,05}{0,1}=0,5M\)
b) mNa2SO4 = 0,05 . 142 = 7,1 g