What volume of 0.540 mol L⁻¹ hydrochloric acid will react completely with 1.34 g of sodium carbonate? A - HSC - SSCE Chemistry - Question 14 - 2023 - Paper 1

The balanced chemical equation for the reaction between sodium carbonate and hydrochloric acid (HCl) is:

$$ext{Na₂CO₃} + 2 ext{HCl} \rightarrow 2 ext{NaCl} + ext{H₂O} + ext{CO₂}$$

From this equation, it can be seen that 1 mole of sodium carbonate reacts with 2 moles of hydrochloric acid.

Next, calculate the moles of hydrochloric acid required:

$$ext{Moles of HCl} = 2 imes ext{Moles of Na₂CO₃} = 2 imes 0.01264 ext{ mol} = 0.02528 ext{ mol}$$

Finally, use the concentration of hydrochloric acid to calculate the volume required. The concentration is 0.540 mol L⁻¹:

ext{Volume (L)} = rac{ ext{Moles of HCl}}{ ext{Concentration}} = rac{0.02528 ext{ mol}}{0.540 ext{ mol/L}} = 0.0468 ext{ L}

Converting this to mL:

$$ext{Volume (mL)} = 0.0468 ext{ L} imes 1000 ext{ mL/L} = 46.8 ext{ mL}$$