Answer:
Step-by-step explanation:
The balanced chemical equation for the reaction between hydrochloric acid (HCl) and sodium hydroxide (NaOH) is:
HCl(aq) + NaOH(s) → NaCl(aq) + H2O(l)
From the balanced equation, we can see that one mole of hydrochloric acid reacts with one mole of sodium hydroxide to produce one mole of sodium chloride and one mole of water.
First, we need to determine which reactant is limiting, i.e., which reactant is completely consumed in the reaction. To do this, we need to compare the number of moles of each reactant, using their respective molar masses:
Molar mass of HCl = 1.008 g/mol (atomic weight of hydrogen) + 35.45 g/mol (atomic weight of chlorine) = 36.46 g/mol
Molar mass of NaOH = 22.99 g/mol (atomic weight of sodium) + 16.00 g/mol (atomic weight of oxygen) + 1.008 g/mol (atomic weight of hydrogen) = 39.99 g/mol
Number of moles of HCl = mass / molar mass = 30.0 g / 36.46 g/mol ≈ 0.823 mol
Number of moles of NaOH = mass / molar mass = 14.3 g / 39.99 g/mol ≈ 0.358 mol
Since the stoichiometric ratio between HCl and NaOH is 1:1, NaOH is the limiting reactant because it has fewer moles than HCl.
Therefore, we can calculate the maximum mass of NaCl that can be produced by the reaction using the number of moles of NaOH:
Number of moles of NaCl produced = number of moles of NaOH used in the reaction = 0.358 mol
Mass of NaCl produced = number of moles of NaCl produced x molar mass of NaCl
Molar mass of NaCl = 22.99 g/mol (atomic weight of sodium) + 35.45 g/mol (atomic weight of chlorine) = 58.44 g/mol
Mass of NaCl produced = 0.358 mol x 58.44 g/mol = 20.9 g
Therefore, the maximum mass of NaCl that can be produced by the reaction is approximately 20.9 g