In the synthesis of K3[Al(C2O4)3].3H2O, 0.0357 moles of the limiting reagent were used in the reaction. What is the maximum possible yield? And which is the limiting reagent?
the answer is 16.5 just figured it out. Thanks anw
H2C2O4.2H2O
To determine the maximum possible yield and the limiting reagent, we need to compare the mole ratio of the reactants with the given moles of the limiting reagent used.
The balanced chemical equation for the synthesis of K3[Al(C2O4)3].3H2O is:
3 K3[Al(C2O4)3].3H2O(aq) → Al2(C2O4)3(s) + 9 K+(aq) + 3 H2O(l)
From the equation, we can see that 1 mole of K3[Al(C2O4)3].3H2O produces 1 mole of Al2(C2O4)3.
Given that 0.0357 moles of the limiting reagent were used, we can conclude that the maximum possible yield of Al2(C2O4)3 is also 0.0357 moles.
Therefore, the maximum possible yield is 0.0357 moles, and the limiting reagent is K3[Al(C2O4)3].3H2O.
To determine the maximum possible yield and the limiting reagent, we need to know the stoichiometry of the reaction. However, since you haven't provided the balanced equation, we can't find the limiting reagent and calculate the maximum possible yield accurately.
To determine the limiting reagent, we usually compare the number of moles of each reactant to the stoichiometric coefficient in the balanced equation. The reactant that has fewer moles compared to its stoichiometric coefficient is the limiting reagent.
Once we know the limiting reagent, we can calculate the maximum possible yield using stoichiometry and the amount of the limiting reagent given.
Please provide the balanced equation for the synthesis reaction, and we can continue from there.