Answer
The volume of carbon dioxide generated from this reaction is equal to 0.106 L.
Work Step by Step
1. Since HCl is a strong acid and it is in excess, the reaction will occur completely.
2. Find the amount of moles of carbon dioxide produced.
$ NaHCO_3 $ : ( 22.99 $\times$ 1 )+ ( 1.008 $\times$ 1 )+ ( 12.01 $\times$ 1 )+ ( 16.00 $\times$ 3 )= 84.01 g/mol
$$ 0.350 \space g \space NaHCO_3 \times \frac{1 \space mole \space NaHCO_3 }{ 84.01 \space g \space NaHCO_3 } \times \frac{ 1 \space mole \space CO_2 }{ 1 \space mole \space NaHCO_3 } = 4.17 \times 10^{-3} \space mole \space CO_2 $$
3. Find the volume using PV = nRT
$$V = \frac{nRT}{P} = \frac{(4.17 \times 10^{-3}) (0.0821) (37.0 + 273.15)}{1.00} $$
$$V =0.106 \space L$$
