Question

6Hydrochloric acid reacts with sodium hydroxide according to the equation:

NaOH + HCl → 2NaCl + H₂O
25 cm of NaOH was neutralized by 26 25 cm of 0 1 mol/dm HCIaCalculate how many moles of HCI was neutralized

Answer 1

Using the molarity of the hydrochloric acid solution (0.1 M) and the volume used in the reaction (25 cm³), it is calculated that 0.0025 moles of hydrochloric acid (HCl) were neutralized by the sodium hydroxide (NaOH).

Step-by-step explanation:

Calculating Moles of HCl Neutralized

To calculate the moles of hydrochloric acid (HCl) that were neutralized by sodium hydroxide (NaOH), we can use the information given and apply the concept of molarity and the balanced chemical equation for the reaction, which shows a 1:1 molar ratio of HCl to NaOH:

HCl(aq) + NaOH(aq) → NaCl(aq) + H₂O(l)

The molarity (M) of a solution is defined as the number of moles of solute (in this case, HCl) per liter of solution. From the titration experiment, we know that 25 cm³ (or 0.025 L) of HCl was used to neutralize the NaOH and that the concentration of HCl is 0.1 mol/dm³ (or 0.1 M).

Using the formula for molarity (M = moles of solute/volume of solution in liters), we can find the number of moles of HCl:

# moles HCl = 0.025 L × 0.1 M = 0.0025 mol HCl

So, 0.0025 moles of HCl was neutralized in the reaction.