How many grams of sodium hydroxide is required to neutralize 73 grams of hydrochloric acid and to from sodium chloride and water
To determine the number of grams of sodium hydroxide (NaOH) required to neutralize hydrochloric acid (HCl) and form sodium chloride (NaCl) and water (H2O), we need to consider the balanced chemical equation for the reaction between NaOH and HCl:
NaOH + HCl → NaCl + H2O
Based on the equation, the stoichiometric ratio between NaOH and HCl is 1:1, meaning that one mole of NaOH reacts with one mole of HCl.
To calculate the number of moles of HCl in 73 grams, we need to know the molar mass of HCl. Hydrogen (H) has an atomic mass of 1 gram/mol, and chlorine (Cl) has an atomic mass of 35.5 grams/mol. So, the molar mass of HCl is 1 + 35.5 = 36.5 grams/mol.
To convert the mass of HCl to moles, we use the formula:
Number of moles = Mass / Molar mass
Substituting the values, we have:
Number of moles of HCl = 73 grams / 36.5 grams/mol ≈ 2 moles
Since the stoichiometric ratio between NaOH and HCl is 1:1, we need 2 moles of NaOH to react with 2 moles of HCl.
To calculate the mass of NaOH required, we use the formula:
Mass = Number of moles × Molar mass
To determine the molar mass of NaOH, sodium (Na) has an atomic mass of 23 grams/mol, oxygen (O) has an atomic mass of 16 grams/mol, and hydrogen (H) has an atomic mass of 1 gram/mol. So, the molar mass of NaOH is 23 + 16 + 1 = 40 grams/mol.
Substituting the values, we have:
Mass of NaOH = 2 moles × 40 grams/mol = 80 grams
Therefore, 80 grams of sodium hydroxide are required to neutralize 73 grams of hydrochloric acid and form sodium chloride and water.