20 g of calcium chloride reacts with 20 g of sodium carbonate to form calcium carbonate and sodium chloride.
CaCl2+ Na2CO3 → CaCO3 + 2NaCl
Calculate the theoretical yield of CaCO3
(Molar mass: Ca 40g/mol; O 16 g/mol; C 12 g/mol; Cl 35.5 g/mol)
First, we need to determine the limiting reactant by using the concept of mole ratios.
The balanced equation tells us that 1 mole of CaCl2 reacts with 1 mole of Na2CO3 to produce 1 mole of CaCO3.
Let's calculate the number of moles of each reactant:
Number of moles of CaCl2 = 20g / 111g/mol = 0.18 mol
Number of moles of Na2CO3 = 20g / 106g/mol = 0.19 mol
Since we have slightly more moles of Na2CO3, it is not the limiting reactant. Therefore, CaCl2 is the limiting reactant.
We can now calculate the theoretical yield of CaCO3 by using the mole ratio:
1 mole of CaCl2 produces 1 mole of CaCO3
0.18 mol of CaCl2 produces 0.18 mol of CaCO3
The molar mass of CaCO3 is:
(40g/mol + 12g/mol + 48g/mol) = 100g/mol
The theoretical yield of CaCO3 is:
0.18 mol x 100g/mol = 18g
Therefore, the theoretical yield of CaCO3 is 18g.
To calculate the theoretical yield of CaCO3, we need to determine the limiting reagent in the reaction. The limiting reagent is the reactant that will be completely consumed and determines the maximum amount of product that can be formed.
First, we need to calculate the number of moles for each reactant:
For CaCl2:
Molar mass of CaCl2 = (Ca: 40 g/mol) + 2(Cl: 35.5 g/mol) = 110 g/mol
Number of moles of CaCl2 = mass / molar mass
Number of moles of CaCl2 = 20 g / 110 g/mol = 0.182 moles
For Na2CO3:
Molar mass of Na2CO3 = (Na: 2 * 23 g/mol) + (C: 12 g/mol) + (O: 3 * 16 g/mol) = 106 g/mol
Number of moles of Na2CO3 = mass / molar mass
Number of moles of Na2CO3 = 20 g / 106 g/mol = 0.189 moles
Now, let's determine the stoichiometric ratios between the reactants and products.
From the balanced equation:
1 mol of CaCl2 produces 1 mol of CaCO3
Comparing the number of moles of CaCl2 and CaCO3, we can see that the stoichiometric ratio is 1:1.
Therefore, the number of moles of CaCO3 formed is also 0.182 moles.
To calculate the mass of CaCO3, we can use the molar mass:
Molar mass of CaCO3 = (Ca: 40 g/mol) + (C: 12 g/mol) + (3 * O: 48 g/mol) = 100 g/mol
Mass of CaCO3 = number of moles of CaCO3 * molar mass
Mass of CaCO3 = 0.182 moles * 100 g/mol = 18.2 g
Therefore, the theoretical yield of CaCO3 is 18.2 grams.