25.0 g of calcium chloride reacts with 25.0 g of sodium carbonate to form calcium carbonate and sodium chloride.
CaCl2+ Na2CO3 → CaCO3 + 2NaCl (Molar mass: Ca 40g/mol; O 16 g/mol; C 12 g/mol; Cl 35.5 g/mol)
Determine the limiting reagent.
In order to determine the limiting reagent, we need to calculate the amount of product that can be formed by each reactant and compare. Let's start with the given masses of both reactants:
- Calcium chloride (CaCl2): 25.0 g / 110.98 g/mol = 0.2250 mol
- Sodium carbonate (Na2CO3): 25.0 g / 105.99 g/mol = 0.2354 mol
Now let's use the balanced equation to calculate the theoretical yield of calcium carbonate (CaCO3) that can be formed by each reactant:
- From 0.2250 mol of CaCl2: 0.2250 mol CaCO3
- From 0.2354 mol of Na2CO3: 0.1177 mol CaCO3 (note: this is half of the theoretical yield since the coefficient of Na2CO3 is 1/2)
Since calcium chloride can produce more calcium carbonate than sodium carbonate, it is the limiting reagent. Therefore, all of the calcium chloride will be consumed in the reaction and some of the sodium carbonate will be left over.
To determine the limiting reagent in a chemical reaction, you need to compare the moles of each reactant to the stoichiometry of the balanced equation.
1. Calculate the number of moles of calcium chloride:
Molar mass of CaCl2 = 40 g/mol + 2(35.5 g/mol) = 111 g/mol
Moles of CaCl2 = (mass of CaCl2) / (molar mass of CaCl2)
= 25.0 g / 111 g/mol
≈ 0.2252 mol
2. Calculate the number of moles of sodium carbonate:
Molar mass of Na2CO3 = 2(23 g/mol) + 12 g/mol + 3(16 g/mol) = 106 g/mol
Moles of Na2CO3 = (mass of Na2CO3) / (molar mass of Na2CO3)
= 25.0 g / 106 g/mol
≈ 0.2358 mol
Based on the stoichiometry of the balanced equation, the mole ratio between CaCl2 and Na2CO3 is 1:1. This means that for every 1 mol of CaCl2, you need 1 mol of Na2CO3 to react completely.
From the calculations, we can see that there are more moles of Na2CO3 (0.2358 mol) compared to CaCl2 (0.2252 mol). Therefore, Na2CO3 is in excess, and CaCl2 is the limiting reagent in this reaction.