Question

In a calorimetry experiment, a student used 80cm³ of 1mHCl and 80cm³ of 1m NaOH in the neutralisation reaction; Calculate the number of moles of water that will be produced.

Answer 1

After using the balanced chemical equation for the reaction between HCl and NaOH, it is determined that 0.080 moles of water are produced when reacting 80cm³ of 1M HCl with 80cm³ of 1M NaOH.

Step-by-step explanation:

To calculate the number of moles of water produced in the neutralization reaction between hydrochloric acid (HCl) and sodium hydroxide (NaOH), we first need to consider the balanced chemical equation:

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

According to the stoichiometry of the reaction, one mole of HCl reacts with one mole of NaOH to produce one mole of water. Given that the student used 80cm³ (which is equivalent to 0.080 L) of 1M HCl and 1M NaOH, we can calculate the moles of each reactant:

Number of moles of HCl = Molarity of HCl × Volume of HCl (in liters) = 1 mol/L × 0.080 L = 0.080 molNumber of moles of NaOH = Molarity of NaOH × Volume of NaOH (in liters) = 1 mol/L × 0.080 L = 0.080 mol

Since the reaction has a 1:1 mole ratio, all of the HCl will react with NaOH to form 0.080 moles of water.