50 mL of a 2.50 M solution of ammonium phosphate is added to 100mL of a 1.50 M solution of calcium nitrate. At the end of the reaction 11.2 grams of precipitate produced. What is the percent yield
To calculate the percent yield, we need to compare the actual yield with the theoretical yield. The theoretical yield is the amount of product that would be obtained if the reaction proceeded perfectly according to the stoichiometry and with 100% efficiency.
First, we need to find the limiting reagent in the reaction. The limiting reagent is the reactant that is completely consumed and determines the maximum amount of product that can be formed. To find the limiting reagent, we compare the number moles of each reactant with the stoichiometric ratio between them.
The balanced chemical equation for the reaction can be written as:
(NH4)3PO4(aq) + 3Ca(NO3)2(aq) → 3Ca(NH4)2PO4(s) + 6NO3^-(aq)
Now, let's calculate the number of moles of each reactant:
Number of moles of ammonium phosphate (NH4)3PO4:
moles = volume (in L) × molarity = 50 mL × (1 L / 1000 mL) × 2.50 M = 0.125 moles
Number of moles of calcium nitrate Ca(NO3)2:
moles = volume (in L) × molarity = 100 mL × (1 L / 1000 mL) × 1.50 M = 0.150 moles
According to the stoichiometry, for every 1 mole of (NH4)3PO4, we need 3 moles of Ca(NO3)2 to completely react. So, the ratio between the number of moles of (NH4)3PO4 and Ca(NO3)2 is 1:3.
Since the ratio of (NH4)3PO4 to Ca(NO3)2 is 1:3, we will need 0.375 moles of Ca(NO3)2 to completely react with 0.125 moles of (NH4)3PO4. However, we only have 0.150 moles of Ca(NO3)2, which is less than what is required. Therefore, Ca(NO3)2 is the limiting reagent.
Now, let's determine the theoretical yield of the precipitate, which is calcium ammonium phosphate (Ca(NH4)2PO4). The molar mass of Ca(NH4)2PO4 is calculated as follows:
Ca: 40.08 g/mol
(NH4)2PO4: (2 × 14.01 g/mol) + (4 × 1.01 g/mol) + 3 × (1.01 g/mol) + 2 × (15.99 g/mol) = 132.17 g/mol
According to the balanced equation, the stoichiometric ratio between Ca(NO3)2 and Ca(NH4)2PO4 is 3:1. Therefore, the moles of Ca(NH4)2PO4 produced will be equal to the moles of Ca(NO3)2 used.
The moles of Ca(NH4)2PO4 produced can be calculated as:
moles = 0.150 moles of Ca(NO3)2
Now, let's calculate the theoretical yield (mass) of the precipitate:
mass = moles × molar mass = 0.150 moles × 132.17 g/mol = 19.83 grams
The theoretical yield of the precipitate is 19.83 grams.
Now, we can calculate the percent yield:
percent yield = (actual yield / theoretical yield) × 100
Given the actual yield is 11.2 grams, the percent yield is:
percent yield = (11.2 grams / 19.83 grams) × 100 ≈ 56.5%
Therefore, the percent yield of the reaction is approximately 56.5%.