Answer:
We can produce 2.35 grams sodium chloride
Explanation:
Step 1: Data given
Mass of hydrochloric acid = 2.2 grams
Molar mass of hydrochloric acid (HCl)= 36.46 g/mol
Mass of sodium hydroxide = 1.61 grams
Molar mass of sodium hydroxide = 40.0 g/mol
Step 2: The balanced equation
HCl + NaOH → NaCl + H2O
Step 3: Calculate moles HCl
Moles HCl = mass HCl / molar mass HCl
Moles HCl = 2.2 grams / 36.46 g/mol
Moles HCl = 0.0603 moles
Step 4: Calculate moles NaOH
Moles NaOH = 1.61 grams / 40.0 g/mol
Moles NaOH = 0.0403 moles
Step 5: Calculate the limiting reactant
NaOH is the limiting reactant. It will completely be consumed (0.0403 moles). HCl is in excess. There will react 0.04025 moles. There will remain 0.0603 -0.0403 = 0.0200 moles
Step 6: Calculate moles sodium chloride
For 1 mol HCl we need 1 mol NaOH to produce 1 mol NaCl and 1 mol H2O
For 0.0403 moles NaOH we'll have 0.0403 moles NaCl
Step 7: Calculate mass NaCl
Mass NaCl = moles NaCl * molar mass NaCl
Mass NaCl = 0.0403 moles * 58.44 g/mol
Mass NaCl = 2.35 grams
We can produce 2.35 grams sodium chloride
