18.5 grams of calcium chloride reacts with 25.7 grams of sodium phosphate and produces calcium phosphate and sodium chloride. What is the limiting reactant? (show work)
I work these problems the long way.
First write and balance the equation.
3CaCl2 + 2Na3PO4 ==> Ca3(PO4)2 + 6NaCl
Convert 18.5g CaCl2 to moles. moles = grams/molar mass. Using the coefficients in the balanced equation, convert mols CaCl2 to moles Ca3(PO4)2. The answer you obtain is the mols Ca3(PO4)2 produced by 18.5 g CaCl2 if you had an excess of the Na3PO4.
Next, do the same process for Na3PO4. This will give you the mols Ca3(PO4)2 produced if you had 25.7 g Na3PO4 and an excess of CaCl2.
The answers for mols Ca3(PO4)2 produced by the two scenarios will not be the same which means one of them is not right; the correct answer in limiting reagent problems is ALWAYS the smaller value and the reagent producing that value is the limiting reagent.
The problem doesn't ask for it but if you want grams produced, then mols Ca3(PO4)2 x molar mass = grams.
To determine the limiting reactant, we need to compare the number of moles of calcium chloride and sodium phosphate.
First, let's calculate the number of moles for each reactant:
Number of moles of calcium chloride:
moles = mass (g) / molar mass (g/mol)
moles = 18.5 g / 110.98 g/mol
moles = 0.1669 mol (rounded to four decimal places)
Number of moles of sodium phosphate:
moles = mass (g) / molar mass (g/mol)
moles = 25.7 g / 163.94 g/mol
moles = 0.1569 mol (rounded to four decimal places)
Next, we need to determine the mole ratio between calcium chloride and sodium phosphate. From the balanced chemical equation, we know that the mole ratio is 1:2. This means that for every 1 mole of calcium chloride, 2 moles of sodium phosphate are required.
Now, we can compare the moles of the two reactants:
Moles of calcium chloride / coefficient = 0.1669 mol / 1 = 0.1669 mol
Moles of sodium phosphate / coefficient = 0.1569 mol / 2 = 0.0785 mol
From the comparison, we can see that the moles of calcium chloride (0.1669 mol) are greater than the moles of sodium phosphate (0.0785 mol). Therefore, the limiting reactant is sodium phosphate because it would be fully consumed before the calcium chloride in the reaction.
Please note that if the moles of calcium chloride were lower than the moles of sodium phosphate, calcium chloride would be the limiting reactant.
To determine the limiting reactant, we need to compare the amounts of calcium chloride and sodium phosphate and find out which one will be completely consumed in the reaction.
Step 1: Write the balanced chemical equation for the reaction:
CaCl2 + Na3PO4 -> Ca3(PO4)2 + 3NaCl
Step 2: Calculate the number of moles for each reactant.
Molar mass of CaCl2 = 40.08 g/mol
Molar mass of Na3PO4 = 164.0 g/mol
Number of moles of CaCl2 = mass of CaCl2 / molar mass of CaCl2
Number of moles of CaCl2 = 18.5 g / 40.08 g/mol ≈ 0.461 moles
Number of moles of Na3PO4 = mass of Na3PO4 / molar mass of Na3PO4
Number of moles of Na3PO4 = 25.7 g / 164.0 g/mol ≈ 0.157 moles
Step 3: Calculate the mole ratio between the two reactants from the balanced equation. In this case, the mole ratio is 1:1 (1 mole of CaCl2 reacts with 1 mole of Na3PO4).
Step 4: Compare the mole ratios of the reactants. The reactant with the smaller mole ratio is the limiting reactant.
Mole ratio of CaCl2:Na3PO4 = 0.461 moles:0.157 moles ≈ 2.93:1
Since the mole ratio of CaCl2:Na3PO4 is approximately 2.93:1, it means there is an excess of CaCl2. Therefore, sodium phosphate (Na3PO4) is the limiting reactant in this reaction.
Note: The stoichiometric ratio plays an important role in determining the limiting reactant by comparing the mole ratios of the given reactants.