A precipitate of calcium phosphate is formed when aqueous solutions of sodium phosphate and excess calcium chloride are mixed. If 100.0 mL of 1.5 M sodium phosphate solution is the limiting reactant, how many grams of calcium phosphate can be expected?
Here is an example of a stoichiometry problem that is worked almost the same as this one. To determine moles, M x L will get it in this problem.
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To determine the number of grams of calcium phosphate that can be expected, we need to use stoichiometry to find the mole ratio between sodium phosphate and calcium phosphate. From the balanced chemical equation for the reaction between sodium phosphate and calcium chloride, we can see that the mole ratio between sodium phosphate and calcium phosphate is 3:1.
Given that we have 100.0 mL of 1.5 M sodium phosphate solution, we need to convert the volume to moles. To do this, we need to know the molar mass of sodium phosphate.
The molar mass of sodium phosphate (Na3PO4) can be calculated as follows:
3(Na) + 1(P) + 4(O) = 3(22.99 g/mol) + 1(30.97 g/mol) + 4(16.00 g/mol) = 163.94 g/mol
Next, we can calculate the number of moles of sodium phosphate:
moles = (concentration in mol/L) * (volume in L)
moles = 1.5 mol/L * (100.0 mL / 1000 mL/L)
moles = 0.15 mol
Since the mole ratio between sodium phosphate and calcium phosphate is 3:1, we can calculate the number of moles of calcium phosphate that can be expected:
moles of calcium phosphate = (moles of sodium phosphate) * (1 mol of calcium phosphate / 3 mol of sodium phosphate)
moles of calcium phosphate = 0.15 mol * (1 mol / 3 mol)
moles of calcium phosphate = 0.05 mol
Finally, we can calculate the mass of calcium phosphate using the molar mass of calcium phosphate (Ca3(PO4)2):
molar mass of calcium phosphate (Ca3(PO4)2) = 3(Ca) + 2(P) + 8(O) = 3(40.08 g/mol) + 2(30.97 g/mol) + 8(16.00 g/mol) = 310.18 g/mol
mass of calcium phosphate = (moles of calcium phosphate) * (molar mass of calcium phosphate)
mass of calcium phosphate = 0.05 mol * 310.18 g/mol
mass of calcium phosphate = 15.509 g
Therefore, approximately 15.509 grams of calcium phosphate can be expected.