how many grams of sodium hydroxide is required to neutralize 73grams of hydrochloric acid and to form sodium chloride and water
To determine the amount of sodium hydroxide needed to neutralize the hydrochloric acid, we need to use the stoichiometry of the balanced chemical equation.
The balanced chemical equation for the reaction between sodium hydroxide (NaOH) and hydrochloric acid (HCl) is as follows:
NaOH + HCl → NaCl + H₂O
From the equation, we can see that the stoichiometric ratio between NaOH and HCl is 1:1. This means that one mole of NaOH reacts with one mole of HCl.
To find the number of moles of HCl in 73 grams, we need to divide the mass (in grams) by the molar mass of HCl. The molar mass of HCl is approximately 36.5 g/mol.
Number of moles of HCl = Mass of HCl / Molar mass of HCl
= 73 g / 36.5 g/mol
≈ 2 moles
Because the stoichiometric ratio between NaOH and HCl is 1:1, we need an equal number of moles of NaOH to neutralize the HCl.
Therefore, the number of moles of NaOH required is also 2.
Now, we need to find the mass of NaOH required for 2 moles. The molar mass of NaOH is approximately 40 g/mol.
Mass of NaOH = Number of moles of NaOH x Molar mass of NaOH
= 2 moles x 40 g/mol
= 80 grams
Therefore, 80 grams of sodium hydroxide is required to neutralize 73 grams of hydrochloric acid and form sodium chloride and water.