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How is mole proportion of synthesis of tetrapotassium iminodisuccinate was performed in three steps: the first step involves the addition of 8 ml of water and 4.4 g of KOH sequentially to 4.6 g of maleic anhydride. second step: 3.6 g ammonium bicarbonate was added gradually stirring for 1 hour at 50°c. the last step: stirring for 0.5 hour at 110 c for the cyclization process.
Accepted Answer
The mole proportion of tetrapotassium iminodisuccinate synthesis can be calculated by considering the stoichiometry of each step. Here's a breakdown:
Step 1:
Maleic anhydride (C4H2O3): 4.6 g / 98 g/mol = 0.047 mol
KOH (potassium hydroxide): 4.4 g / 56.11 g/mol = 0.078 mol
The reaction is likely a ring-opening reaction of maleic anhydride with KOH, resulting in the formation of a potassium salt of maleic acid. The stoichiometry suggests a 1:2 mole ratio of maleic anhydride to KOH.
Step 2:
Ammonium bicarbonate (NH4HCO3): 3.6 g / 79.07 g/mol = 0.045 mol
Ammonium bicarbonate likely reacts with the potassium salt of maleic acid to form the iminodisuccinate anion. The exact stoichiometry depends on the reaction mechanism.
Step 3:
The cyclization step involves the formation of the cyclic iminodisuccinate ring. It's difficult to calculate the exact mole proportion without knowing the specific reaction mechanism and product yield.
Overall:
To determine the mole proportion of tetrapotassium iminodisuccinate, you need to know the yield of each step and the exact stoichiometry of the reactions involved. Further investigation is needed to fully analyze the mole proportions.
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