Sodium phosphate dodecahydate is reacts with barium chloride dihydrate.barium phosphate is lost during filtering will the percent limiting reactant be hihg or low?
To determine whether the percent limiting reactant will be high or low, we first need to identify the limiting reactant in the reaction. The limiting reactant is the reactant that is completely consumed and limits the amount of product that can be formed.
The balanced chemical equation for the reaction between sodium phosphate dodecahydrate (Na3PO4 · 12H2O) and barium chloride dihydrate (BaCl2 · 2H2O) is:
2Na3PO4 · 12H2O + 3BaCl2 · 2H2O → Ba3(PO4)2 + 6NaCl + 34H2O
From the equation, we can see that 2 moles of Na3PO4 · 12H2O react with 3 moles of BaCl2 · 2H2O to produce 1 mole of Ba3(PO4)2.
To determine the limiting reactant, we need to compare the amount of each reactant available relative to their stoichiometric coefficients. Let's assume we have 2 moles of Na3PO4 · 12H2O and 3 moles of BaCl2 · 2H2O.
The molar ratio between Na3PO4 · 12H2O and Ba3(PO4)2 is 2:1. Therefore, 2 moles of Na3PO4 · 12H2O would require 1 mole of BaCl2 · 2H2O to react completely.
Since there are 3 moles of BaCl2 · 2H2O available, it is in excess compared to the 2 moles of Na3PO4 · 12H2O. Thus, Na3PO4 · 12H2O is the limiting reactant.
Now, let's consider the loss of barium phosphate during filtering. If barium phosphate is lost during the filtering process, it means that some of the product is not retained. As a result, the actual yield of Ba3(PO4)2 will be lower than the theoretical yield.
Since the limiting reactant determines the amount of product that can be formed, the loss of barium phosphate will not affect its concentration or the amount of limiting reactant consumed. However, it will decrease the actual yield of Ba3(PO4)2.
Therefore, the percent limiting reactant (Na3PO4 · 12H2O) will be high because it is fully consumed, but the actual yield of the product (Ba3(PO4)2) will be lower due to the loss during filtering.