Volume of ${\text{CO}}_2$  at STP formed when 100 ml of 0.1M ${\text{H}}_2{\text{SO}}_4$  reacts with excess of ${\text{Na}}_2{\text{CO}}_3$ is

Volume of ${\text{H}}_2{\text{SO}}_4$ ,$\text{V}=100\text{ mL}$  

Since $1000\text{mL}=1\text{L}$

So, $\frac{1\text{L}}{1000\text{mL}}\times 100\text{mL=}\text{0}\text{.1}\text{L}$

Molarity of ${\text{H}}_2{\text{SO}}_4$  solution$=0.1\text{ M}$

Number of moles of ${\text{H}}_2{\text{SO}}_4$  ,$n=?$

From molarity equation,

             $\text{M}=\frac{n}{\text{V}\text{in liters}}$

So, $\text{n}=\text{M}\times \text{V}$

Then, $n=0.1\times 0.1\text{=}\text{0}\text{.01}\text{mol}$                                  

The balanced equation,                                    

${\text{Na}}_2{\text{CO}}_3+{\text{H}}_2{\text{SO}}_4\to {\text{Na}}_2{\text{SO}}_4+{\text{H}}_2\text{O}+{\text{CO}}_2$

1 mole ${\text{H}}_2{\text{SO}}_4$  produce 22.4 lit of ${\text{CO}}_2$

0.01 mole ${\text{H}}_2{\text{SO}}_4$  produce $0.224\text{ L}$ .