Which of the following is/are correct?

The following reaction occurs:

${\mathrm{Na}}_2{\mathrm{CO}}_3+2\mathrm{HCl}⟶2\mathrm{NaCl}+{\mathrm{CO}}_2+{\mathrm{H}}_2\mathrm{O}$

106.0 g of Na₂CO₃ reacts with 109.5 g of HCl.

$\left(\mathrm{Mw}\text{ of }{\mathrm{Na}}_2{\mathrm{CO}}_3=106,\mathrm{Mw}\text{ of }\mathrm{HCl}=36.5,\mathrm{Mw}\text{ of }\mathrm{NaCl}=58.5\right)$

$\mathrm{Moles} \mathrm{of} {\mathrm{Na}}_2{\mathrm{CO}}_3=\frac{106}{106}=1.0 \mathrm{mol}$

Moles of HCl = $\frac{109.5}{36.5}=3.0 \mathrm{mol}$

1. Since for 1 mol of ${\mathrm{Na}}_2{\mathrm{CO}}_3,2 \mathrm{mol} \mathrm{of} \mathrm{HCl} \mathrm{is} \mathrm{required}. \mathrm{So}, \mathrm{HCl} \mathrm{is} \mathrm{in} \mathrm{excess} (3-2) = 1.0 \mathrm{mol}$

Therefore, Na₂CO₃ is the limiting quantity.

2. Weight of NaCl formed 

$\begin{array}{l}=\left(1.0 \mathrm{mol} {\mathrm{Na}}_2{\mathrm{CO}}_3\right)\left(\frac{2 \mathrm{mol} \mathrm{NaCl}}{\mathrm{mol} {\mathrm{Na}}_2{\mathrm{CO}}_3}\right)\left(\frac{58.5 \mathrm{g} \mathrm{NaCl}}{\mathrm{mol} \mathrm{NaCl}}\right)\\ =1\times 2\times 58.5=117.0 \mathrm{g} \mathrm{NaCl}\end{array}$

=1 x 2 x 58.5 = 117.0 g NaCl

3. 1 mol of Na₂CO₃ = 1 mol of CO₂ = 22.7 L at 1 bar, 273 K

4. 1 mol of Na₂CO₃ = 1 mol of CO₂ = 24.7 L at 1 bar, 298 K