How many mL of 0.098 M HCl is required to completely react with 30.0 mL of 0.150 M K2CO3 solution according to the equation below? What is the molarity of KCl solution formed?                                    2HCl(aq) + K2CO3(aq) ⟶ 2KCl(aq) + H2O(l) + CO2(g)

It is given that 30.0 mL of 0.150 M K₂CO₃ is reacting completely with 0.098 M of HCl and the reaction is shown below,

$2{\mathrm{HCl}}_{\left(\mathrm{aq}\right)} + {\mathrm{K}}_2{\mathrm{CO}}_{3\left(\mathrm{aq}\right)} ⟶ 2{\mathrm{KCl}}_{\left(\mathrm{aq}\right)} + {\mathrm{H}}_2{\mathrm{O}}_{\left(\mathrm{l}\right)} + {\mathrm{CO}}_{2\left(\mathrm{g}\right)}$

To calculate the moles of K₂CO₃ present,

$$\begin{gathered} \mathrm{no}. \mathrm{of} \mathrm{moles} = \mathrm{molarity} \times \mathrm{volume} \mathrm{in} \mathrm{litres} \\ \mathrm{moles} \mathrm{of} {\mathrm{K}}_2{\mathrm{CO}}_3= 0.150 \mathrm{M} \times 0.030 \mathrm{L} \\ = 0.0045 \mathrm{moles} \end{gathered}$$

To calculate the moles of HCl present with the help of the balanced equation,

$$\begin{gathered} \mathrm{In} \mathrm{the} \mathrm{balanced} \mathrm{reaction} \mathrm{as} 1 \mathrm{mole} \mathrm{of} {\mathrm{K}}_2{\mathrm{CO}}_3 \mathrm{is} \mathrm{reacting} \mathrm{with} 2 \mathrm{moles} \mathrm{of} \mathrm{HCl}, \\ \mathrm{therefore}, 0.0045 \mathrm{moles} \mathrm{of} {\mathrm{K}}_2{\mathrm{CO}}_3 \mathrm{will} \mathrm{react} \mathrm{with} 2\times 0.0045 = 0.009 \mathrm{moles} \mathrm{of} \mathrm{HCl}. \end{gathered}$$