3.2 grams of calcium chloride reacts with 3.55 grams of sodium sulfate producing calcium sulfate + sodium chloride. Which is the reactant in excess and the excess mass?
Cacl2 + Na2SO4 = CaSO4 + NaCl
A slight twist on limiting reagent (LR).
CaCl2 _+ Na2SO4 ==> CaSO4 + 2NaCl
mols CaCl2 = grams/molar mass = ?
mols Na2SO4 = grams/molar mass = ?
Using the coefficients in the balanced equationk convert mols Cacl2 to mols CaSO4.
Do the same and convert mols Na2SO4 to mols CaSO4.
It is likely that these two values will not agree; the correct value in LR problems is ALWAYS the smaller value the reagent responsible for that value is the LR. The other is the excess reagent (ER).
To find how much of the ER is used, convert mols LR to mols ER. Subtract the initial amount of the ER - amount ER used = amount ER remaining unreacted.
To identify the reactant in excess, we need to compare the stoichiometry of the reactants and their actual masses used in the reaction.
Let's start by calculating the number of moles for each reactant using their molar masses:
Molar mass of CaCl2:
- Calcium (Ca) = 40.08 g/mol
- Chlorine (Cl) = 35.45 g/mol (x2, since there are two chlorine atoms in CaCl2)
Molar mass of CaCl2 = 40.08 + (35.45 x 2) = 40.08 + 70.90 = 111.98 g/mol
Molar mass of Na2SO4:
- Sodium (Na) = 22.99 g/mol (x2, since there are two sodium atoms in Na2SO4)
- Sulfur (S) = 32.07 g/mol
- Oxygen (O) = 16.00 g/mol (x4, since there are four oxygen atoms in Na2SO4)
Molar mass of Na2SO4 = (22.99 x 2) + 32.07 + (16.00 x 4) = 45.98 + 32.07 + 64.00 = 142.05 g/mol
Next, we can calculate the moles of each reactant used in the reaction:
Moles of CaCl2 = mass of CaCl2 / molar mass of CaCl2
Moles of CaCl2 = 3.2 g / 111.98 g/mol ≈ 0.0286 mol
Moles of Na2SO4 = mass of Na2SO4 / molar mass of Na2SO4
Moles of Na2SO4 = 3.55 g / 142.05 g/mol ≈ 0.0250 mol
By examining the balanced chemical equation:
CaCl2 + Na2SO4 = CaSO4 + 2NaCl
We can see that 1 mole of CaCl2 reacts with 1 mole of Na2SO4. Therefore, the stoichiometric ratio between CaCl2 and Na2SO4 is 1:1.
Since the moles of Na2SO4 (0.0250 mol) are less than the moles of CaCl2 (0.0286 mol) in the actual reaction, Na2SO4 is the reactant that is consumed in its entirety.
Thus, CaCl2 is the reactant in excess.
To determine the excess mass of CaCl2, we need to find the remaining moles of CaCl2 after the reaction consumes all of the Na2SO4:
Remaining moles of CaCl2 = Moles of CaCl2 - Moles of Na2SO4
Remaining moles of CaCl2 = 0.0286 mol - 0.0250 mol = 0.0036 mol
The excess mass of CaCl2 can be calculated by multiplying the remaining moles by the molar mass of CaCl2:
Excess mass of CaCl2 = Remaining moles of CaCl2 x Molar mass of CaCl2
Excess mass of CaCl2 = 0.0036 mol x 111.98 g/mol ≈ 0.4037 g
Therefore, the reactant in excess is CaCl2 and the excess mass of CaCl2 is approximately 0.4037 grams.