which reactant in the reaction of sodium carbonate and calcium chloride dihydrate is the limiting reactant if 1.00 gram of each reagent is used?
a. sodium carbonate
b. calcium chloride dihydrate
Most limiting reagent problems are done the same way.
1. Write the equation and balance it.
2a. Convert 1 g Na2CO3 to moles. moles = grams/molar mass.
2b. Convert 1 g CaCl2.2H2O to moles. same procedure.
3a. Using the coefficients in the balanced equation, convert moles Na2CO3 to moles CaCO3.
3b. Same procedure, convert moles CaCl2.2H2O to moles CaCO3.
3c. It is quite likely that moles from 3a and 3b will not be the same which means one of them is wrong. In limiting reagent problems, the correct answer is ALWAYS the smaller value and the reagent producing that value is the limiting reagent.
4. You aren't asked to calculate grams of the product, but you can convert the value from 3c to grams. g = moles x molar mass.
To determine the limiting reactant in a chemical reaction, you need to compare the number of moles of each reactant used. This requires converting the mass of each reactant (in this case 1.00 gram of each) to moles.
Let's start by calculating the number of moles for each reactant:
1. Sodium Carbonate (Na2CO3):
- The molar mass of Na2CO3 is 105.99 g/mol (sodium: 22.99 g/mol, carbon: 12.01 g/mol, oxygen: 16.00 g/mol).
- Divide the given mass (1.00 gram) by the molar mass to obtain the number of moles: 1.00 g / 105.99 g/mol = 0.00943 mol.
2. Calcium Chloride Dihydrate (CaCl2·2H2O):
- The molar mass of CaCl2·2H2O is 147.02 g/mol (calcium: 40.08 g/mol, chlorine: 35.45 g/mol, hydrogen: 1.01 g/mol, oxygen: 16.00 g/mol).
- Divide the given mass (1.00 gram) by the molar mass to obtain the number of moles: 1.00 g / 147.02 g/mol = 0.00680 mol.
Next, compare the number of moles of each reactant used. In this case, sodium carbonate has 0.00943 mol, and calcium chloride dihydrate has 0.00680 mol.
Since the reaction requires a 1:1 ratio between sodium carbonate and calcium chloride dihydrate, the limiting reactant is the reactant with fewer moles. In this case, the calcium chloride dihydrate (CaCl2·2H2O) is the limiting reactant because it has fewer moles compared to sodium carbonate (Na2CO3).
Therefore, the answer is b. Calcium Chloride Dihydrate.