2M+2H2O\(\rightarrow\)2MOH+H2
2MOH+H2SO4\(\rightarrow\)M2SO4+2H2O
\(n_{H_2}=\dfrac{0,336}{22,4}=0,015mol\)
\(n_{MOH}=2n_{H_2}=2.0,015=0,03mol\)
\(n_{H_2SO_4}=\dfrac{1}{2}n_{NaOH}=\dfrac{0,03}{2}=0,015mol\)
\(m_{H_2SO_4}=0,015.98=1,47gam\)
\(m_{dd_{H_2SO_4}}=\dfrac{1,47.100}{2,94}=50gam\)