Answer:
Explanation:
Initial burette reading = 1.81 mL
final burette reading = 39.7 mL
volume of NaOH used = 39.7 - 1.81 = 37.89 mL .
37.89 mL of .1029 M NaOH is used to neutralise triprotic acid
No of moles contained by 37.89 mL of .1029 M NaOH
= .03789 x .1029 moles
= 3.89 x 10⁻³ moles
Since acid is triprotic , its equivalent weight = molecular weight / 3
No of moles of triprotic acid = 3.89 x 10⁻³ / 3
= 1.30 x 10⁻³ moles .
Answer:
1.772 gram is the approximate answer
Explanation:
molecular mass of AlCl3 is 132 g per mole and of Al(OH)3 is 78 g per mole
the reaction is
AlCl3 + 3 NaOH ---> Al(OH)3 + 3 NaCl
from the reaction it is clear that 1 mole AlCl3 makes 1 mole Al(OH)3
implies 132g AlCl3 gives 78g Al(OH)3
Implies 3g AlCl3 gives
3*122/78 = 1.772 grams
