How many grams of sodium carbonate is required for a complete reaction with 1.00g of calcium chloride dihydrate? Help!
To determine the amount of sodium carbonate required for a complete reaction with 1.00g of calcium chloride dihydrate, you need to consider the stoichiometry of the reaction.
The balanced chemical equation for the reaction between sodium carbonate (Na2CO3) and calcium chloride dihydrate (CaCl2·2H2O) is:
Na2CO3 + CaCl2·2H2O → 2NaCl + CaCO3 + 2H2O
From the equation, you can see that for every 1 mole of calcium chloride dihydrate, you need 1 mole of sodium carbonate to achieve a complete reaction.
1. Calculate the molar mass of calcium chloride dihydrate (CaCl2·2H2O):
- Molar mass of Ca = 40.08 g/mol
- Molar mass of Cl = 35.45 g/mol (x2 since there are two Cl atoms)
- Molar mass of H = 1.01 g/mol (x2 since there are two H atoms)
- Molar mass of O = 16.00 g/mol (x2 since there are two O atoms)
- Molar mass of CaCl2·2H2O = 40.08 g/mol + (35.45 g/mol x 2) + (1.01 g/mol x 2) + (16.00 g/mol x 2)
- Molar mass of CaCl2·2H2O = 147.02 g/mol
2. Convert the mass of calcium chloride dihydrate to moles:
- Moles of CaCl2·2H2O = Mass (g) / Molar mass (g/mol)
- Moles of CaCl2·2H2O = 1.00 g / 147.02 g/mol
- Moles of CaCl2·2H2O ≈ 0.0068 mol
3. Since the reaction is a 1:1 ratio, the moles of sodium carbonate required will be the same as the moles of calcium chloride dihydrate. Therefore, the answer is:
- Moles of Na2CO3 required = Moles of CaCl2·2H2O
- Moles of Na2CO3 required ≈ 0.0068 mol
4. Finally, convert moles of sodium carbonate to grams:
- Mass (g) of Na2CO3 required = Moles of Na2CO3 x Molar mass of Na2CO3
- Mass (g) of Na2CO3 required ≈ 0.0068 mol x [(22.99 g/mol x 2) + 12.01 g/mol + (16.00 g/mol x 3)]
- Mass (g) of Na2CO3 required ≈ 0.0068 mol x 105.99 g/mol
- Mass (g) of Na2CO3 required ≈ 0.724 g
So, approximately 0.724 grams of sodium carbonate is required for a complete reaction with 1.00 gram of calcium chloride dihydrate.