Question: "Calculate the theoretical yield of calcium carbonate of you react 2.55g of calcium chloride dihydrate with an excess amount of sodium carbonate."
I got the molar mass of calcium chloride dihydrate which is 131.015 g/mol, but Im not sure how to start the equation for the theoretical yield...
2.55g X ?????
I would use percentage of calcium in CaCl2.2H2O
Percent= molmassCa/131.015
muliply that by 2.55..
you would use Calcium because its the limiting reactant?
So... 40.078/131.015 = .305903
Then... 2.55g X .305903g Ca/1g = the theoretical yield?
To calculate the theoretical yield of calcium carbonate, you need to use stoichiometry. Stoichiometry is a mathematical tool used to determine the relationships between reactants and products in a chemical reaction. In this case, the balanced chemical equation for the reaction between calcium chloride dihydrate (CaCl2·2H2O) and sodium carbonate (Na2CO3) is needed.
The balanced chemical equation for the reaction is:
CaCl2·2H2O + Na2CO3 → CaCO3 + 2NaCl + 2H2O
From the equation, you can see that 1 mole of CaCl2·2H2O reacts to produce 1 mole of CaCO3. Therefore, the molar ratio between CaCl2·2H2O and CaCO3 is 1:1.
To calculate the moles of CaCl2·2H2O, you can use the molar mass you found (131.015 g/mol) and the given mass of 2.55 g. Divide the given mass by the molar mass:
moles of CaCl2·2H2O = 2.55 g / 131.015 g/mol
Now, since the stoichiometric ratio is 1:1 between CaCl2·2H2O and CaCO3, the moles of CaCO3 will be the same as the moles of CaCl2·2H2O.
Therefore, the theoretical yield of calcium carbonate can be calculated by multiplying the moles of CaCl2·2H2O by the molar mass of CaCO3 (100.086 g/mol):
theoretical yield of CaCO3 = moles of CaCl2·2H2O × molar mass of CaCO3
So, the equation for the theoretical yield is:
2.55 g × (1 mol / 131.015 g/mol) × (100.086 g/mol / 1 mol)
Simplifying the equation gives:
theoretical yield of CaCO3 = 2.55 g × 100.086 g/131.015 g
Now, calculate the value using the given equation to find the theoretical yield of calcium carbonate.