Calculating percentage yield of *sodium chloride
Seriously stuck with this, any help would be much appreciated
How on earth do i calculate percentage yield from this practical in preparing sodium chloride
-25cm3 of hydrochloric acid (standard solution)
-Burette filled with sodium hydroxide (alkali solution)
Titration:
Trial 1: 0 - 23.8
Trial 2: 0 -20.6
Trial 3: 0 - 21.6
Average titre = 22
After crystallisation
Mass of weighing boat – 1.29g
Mass of weighing boat + sodium chloride – 1.41g
Mass of sodium chloride – 0.12g
Word equation:
Sodium hydroxide + hydrochloric acid -> sodium chloride + water NaOH + HCl -> NaCl + H2O
Percentage yield:
Molecular mass of sodium hydroxide – 40.00g/mol (2dp)
Molecular mass of hydrochloric acid – 36.46g/mol (2dp)
How on earth do i calculate percentage yield from this practical in preparing sodium chloride
-25cm3 of hydrochloric acid (standard solution)
-Burette filled with sodium hydroxide (alkali solution)
Titration:
Trial 1: 0 - 23.8
Trial 2: 0 -20.6
Trial 3: 0 - 21.6
Average titre = 22
After crystallisation
Mass of weighing boat – 1.29g
Mass of weighing boat + sodium chloride – 1.41g
Mass of sodium chloride – 0.12g
Word equation:
Sodium hydroxide + hydrochloric acid -> sodium chloride + water NaOH + HCl -> NaCl + H2O
Percentage yield:
Molecular mass of sodium hydroxide – 40.00g/mol (2dp)
Molecular mass of hydrochloric acid – 36.46g/mol (2dp)
Original post by Adam__nicholson
Seriously stuck with this, any help would be much appreciated
How on earth do i calculate percentage yield from this practical in preparing sodium chloride
-25cm3 of hydrochloric acid (standard solution)
-Burette filled with sodium hydroxide (alkali solution)
Titration:
Trial 1: 0 - 23.8
Trial 2: 0 -20.6
Trial 3: 0 - 21.6
Average titre = 22
After crystallisation
Mass of weighing boat – 1.29g
Mass of weighing boat + sodium chloride – 1.41g
Mass of sodium chloride – 0.12g
Word equation:
Sodium hydroxide + hydrochloric acid -> sodium chloride + water NaOH + HCl -> NaCl + H2O
Percentage yield:
Molecular mass of sodium hydroxide – 40.00g/mol (2dp)
Molecular mass of hydrochloric acid – 36.46g/mol (2dp)
How on earth do i calculate percentage yield from this practical in preparing sodium chloride
-25cm3 of hydrochloric acid (standard solution)
-Burette filled with sodium hydroxide (alkali solution)
Titration:
Trial 1: 0 - 23.8
Trial 2: 0 -20.6
Trial 3: 0 - 21.6
Average titre = 22
After crystallisation
Mass of weighing boat – 1.29g
Mass of weighing boat + sodium chloride – 1.41g
Mass of sodium chloride – 0.12g
Word equation:
Sodium hydroxide + hydrochloric acid -> sodium chloride + water NaOH + HCl -> NaCl + H2O
Percentage yield:
Molecular mass of sodium hydroxide – 40.00g/mol (2dp)
Molecular mass of hydrochloric acid – 36.46g/mol (2dp)
If you know your concentration of HCl, multiply concentration by volume (dm3) to get moles.
As this reacts on a 1:1 ratio, this is also the munber of moles of NaOH.
This reaction would also produce the same number of moles of NaCl.
Multiply the number of moles of NaCl by its molecular mass to find the mass (should be 0.12g)
Get the number of moles of H2O (same as used for everything else) x18g/mol to get mass of water.
Mass of NaCl divided by (mass of NaCl plus mass of H2O), x100 gives percentage yield.
Hope this helps.
Life saver, thank you so much
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