You mix 50 mL of 0.1 M ammonium phosphate with 75 mL of 0.05 M calcium nitrate. How much of a precipitate forms from this reaction (in grams)?
2(NH4)3PO4 + 3Ca(NO3)2 ==> Ca3(PO4)2 + 6NH4NO3
moles (NH4)3PO4 = M x L
moles Ca(NO3)2 = M x L.
Using the coefficients in the balanced equation, convert moles (NH4)3PO4 to moles Ca3(PO4)2.
Same procedure, convert moles Ca(NO3)2 to moles Ca3(PO4)2.
In limiting reagent problems, the answers do NOT agree; therefore, one of them must be wrong. In limiting reagent problems, the correct value is ALWAYS the smaller value and the reagent producing that value is the limiting reagent.
Using the smaller value, convert moles to grams. g = moles x molar mass.
To determine the amount of precipitate formed in this reaction, we need to first identify the balanced equation for the reaction between ammonium phosphate (NH4)3PO4 and calcium nitrate Ca(NO3)2.
The balanced chemical equation for the reaction between these two compounds is:
3(NH4)3PO4 + 2Ca(NO3)2 → Ca3(PO4)2 + 6NH4NO3
From the balanced equation, we can see that 3 moles of ammonium phosphate react with 2 moles of calcium nitrate to produce 1 mole of calcium phosphate (Ca3(PO4)2) and 6 moles of ammonium nitrate (NH4NO3).
Now, let's calculate the number of moles of each reactant present in the solution:
For ammonium phosphate:
Molarity of ammonium phosphate = 0.1 M
Volume of ammonium phosphate = 50 mL = 0.05 L
Number of moles of ammonium phosphate = Molarity × Volume
= 0.1 M × 0.05 L
= 0.005 moles
For calcium nitrate:
Molarity of calcium nitrate = 0.05 M
Volume of calcium nitrate = 75 mL = 0.075 L
Number of moles of calcium nitrate = Molarity × Volume
= 0.05 M × 0.075 L
= 0.00375 moles
Next, we need to determine the limiting reactant, which is the reactant that is completely consumed and determines the maximum amount of product that can be formed.
To find the limiting reactant, we compare the number of moles of each reactant with the stoichiometric ratio from the balanced equation.
According to the balanced equation, the stoichiometric ratio between ammonium phosphate and calcium nitrate is 3:2.
Moles of ammonium phosphate / Stoichiometric ratio of ammonium phosphate = 0.005 moles / 3 = 0.00167 moles
Moles of calcium nitrate / Stoichiometric ratio of calcium nitrate = 0.00375 moles / 2 = 0.001875 moles
Since 0.00167 moles of ammonium phosphate is less than 0.001875 moles of calcium nitrate, ammonium phosphate is the limiting reactant.
From the balanced equation, we know that 3 moles of ammonium phosphate react to form 1 mole of calcium phosphate. Therefore, 0.00167 moles of ammonium phosphate will produce:
0.00167 moles x (1 mole Ca3(PO4)2 / 3 moles (NH4)3PO4) = 0.0005567 moles of calcium phosphate
Finally, we can calculate the mass of calcium phosphate using its molar mass:
Molar mass of calcium phosphate (Ca3(PO4)2) = 310.18 g/mol
Mass of calcium phosphate = Moles of calcium phosphate × Molar mass
= 0.0005567 moles × 310.18 g/mol
= 0.172 g
Therefore, approximately 0.172 grams of calcium phosphate will form as a precipitate in this reaction.