Consider the reaction represented by the following equation: Na2CO3(aq) + 2HCl(aq) \(\longrightarrow\) 2NaCl(aq) + H2O(l) + CO2(g) . What volume of 0.02 mol...

The balanced chemical equation for the reaction between Na₂CO_3(aq) and HCl_(aq) is: Na₂CO_3(aq) + 2HCl_(aq) → 2NaCl_(aq) + H₂O_(l) + CO_2(g) From the equation, we can see that 1 mole of Na₂CO_3(aq) reacts with 2 moles of HCl_(aq) to produce 2 moles of NaCl_(aq), 1 mole of H₂O_(l), and 1 mole of CO_2(g). Therefore, the number of moles of HCl_(aq) that reacted can be calculated as: number of moles = concentration x volume number of moles = 0.10 mol dm^-3 x 0.040 dm³ number of moles = 0.004 mol From the balanced equation, we know that 1 mole of Na₂CO_3(aq) reacts with 2 moles of HCl_(aq). Therefore, the number of moles of Na₂CO_3(aq) required to completely neutralize the HCl_(aq) can be calculated as: number of moles = 0.5 x number of moles of HCl_(aq) number of moles = 0.5 x 0.004 number of moles = 0.002 Finally, we can use the concentration and number of moles to calculate the required volume of Na₂CO_3(aq) as: volume = number of moles / concentration volume = 0.002 mol / 0.02 mol dm^-3 volume = 0.1 dm³ = 100 cm³ Therefore, the correct answer is 100 cm³.