What mass of  ${\mathrm{NaHCO}}_3 (\mathrm{M}=84,0)$ is required to completely neutralize 25.0 mL of  $0.125 \mathrm{M} {\mathrm{H}}_2{\mathrm{SO}}_4$ ?

The reaction takes place as,

$2{\mathrm{NaH}}_3+{\mathrm{H}}_2{\mathrm{S}}_4\to {\mathrm{Na}}_2{\mathrm{s}}_4+2{\mathrm{H}}_2\mathrm{O}+2{\mathrm{cO}}_2$

mass required to completely neutralize 25.0 mL of 0.125 M 

$\frac{25\times 0.125\times 2}{1000}\times 84$