50.0ml of an aqueous solution of 2.50M H3PO4 are mixed with 75.0 ml of an aqueous solution of 4.00 M Ca(OH)2. The resulting precipitate is washed and isolated using filtration. What mass of Ca3(PO4)2 is obtained?
This is a limiting reagent (LR) problem. I know that becsuer amounts are given for BOTH reactants.
2H3PO4 + 3Ca(OH)2 ==> Ca3(PO4)2 + 6H2O
1. mols H3PO4 = M x L = ?
Do the same for mols Ca(OH)2.
2a. Using the coefficients in the balanced equation, convert mols H3PO4 to mols Ca3(PO4)2.
2b. Do the same for converting mols Ca(OH)2 to mols Ca3(PO4)2.
2c. It is likely that the values in 2a and 2b will not agree which means one of them is not right. The correct value in LR problems is ALWAYS the smaller value and the reagent producing that value is called the LR.
3. Now using the smaller value for mols,convert to grams. g = mols x molar mass.
To find the mass of Ca3(PO4)2 obtained, we need to determine the limiting reactant in the reaction between H3PO4 and Ca(OH)2. Once we know the limiting reactant, we can calculate the amount of Ca3(PO4)2 formed using stoichiometry.
First, let's determine the number of moles of H3PO4 and Ca(OH)2 in each solution:
Volume of H3PO4 solution = 50.0 ml = 0.050 L
Concentration of H3PO4 solution = 2.50 M
Number of moles of H3PO4 = concentration × volume = 2.50 M × 0.050 L = 0.125 moles
Volume of Ca(OH)2 solution = 75.0 ml = 0.075 L
Concentration of Ca(OH)2 solution = 4.00 M
Number of moles of Ca(OH)2 = concentration × volume = 4.00 M × 0.075 L = 0.300 moles
Now, let's determine the stoichiometric ratio between H3PO4 and Ca3(PO4)2. From the balanced chemical equation:
2 H3PO4 + 3 Ca(OH)2 → Ca3(PO4)2 + 6 H2O
We can see that the stoichiometric ratio is 2:1 between H3PO4 and Ca3(PO4)2. This means that for every 2 moles of H3PO4, we should obtain 1 mole of Ca3(PO4)2.
Next, we need to determine the limiting reactant. The limiting reactant is the one that is fully consumed and determines the maximum amount of product that can be formed.
From the stoichiometric ratio, we can determine the number of moles of Ca3(PO4)2 that can be formed from H3PO4:
Number of moles of Ca3(PO4)2 = 0.125 moles of H3PO4 × (1 mole of Ca3(PO4)2 / 2 moles of H3PO4) = 0.0625 moles
From the stoichiometric ratio, we can also determine the number of moles of Ca3(PO4)2 that can be formed from Ca(OH)2:
Number of moles of Ca3(PO4)2 = 0.300 moles of Ca(OH)2 × (1 mole of Ca3(PO4)2 / 3 moles of Ca(OH)2) = 0.100 moles
Since the number of moles of Ca3(PO4)2 formed is determined by the limiting reactant, the limiting reactant is H3PO4, as it forms less Ca3(PO4)2 compared to Ca(OH)2.
Finally, to calculate the mass of Ca3(PO4)2 formed, we need to use its molar mass. The molar mass of Ca3(PO4)2 can be calculated by adding up the atomic masses of its constituent elements:
Molar mass of Ca = 40.08 g/mol
Molar mass of P = 30.97 g/mol
Molar mass of O = 16.00 g/mol
Molar mass of Ca3(PO4)2 = (3 × 40.08) + (2 × 30.97) + (8 × 16.00) = 310.18 g/mol
Finally, let's calculate the mass of Ca3(PO4)2:
Mass of Ca3(PO4)2 = Number of moles of Ca3(PO4)2 × Molar mass of Ca3(PO4)2
= 0.0625 moles × 310.18 g/mol
= 19.38 g
Therefore, the mass of Ca3(PO4)2 obtained is 19.38 grams.