How many grams of sodium carbonate is required for a complete reaction with 1.00g of calcium chloride dihydrate? Help!
To determine the number of grams of sodium carbonate required for a complete reaction with 1.00g of calcium chloride dihydrate, we need to calculate the molar mass of both compounds and use stoichiometry.
1. First, let's calculate the molar mass of calcium chloride dihydrate (CaCl2·2H2O):
- Molar mass of calcium (Ca) = 40.08 g/mol
- Molar mass of chlorine (Cl) = 35.45 g/mol
- Molar mass of water (H2O) = 18.02 g/mol
Since calcium chloride dihydrate has two chlorine atoms and two water molecules, we can calculate the molar mass as follows:
Molar mass of CaCl2·2H2O = (40.08 g/mol) + (2 × 35.45 g/mol) + (2 × 18.02 g/mol)
2. Calculate the molar mass of sodium carbonate (Na2CO3):
- Molar mass of sodium (Na) = 22.99 g/mol
- Molar mass of carbon (C) = 12.01 g/mol
- Molar mass of oxygen (O) = 16.00 g/mol
Therefore, the molar mass of sodium carbonate is:
Molar mass of Na2CO3 = (2 × 22.99 g/mol) + (1 × 12.01 g/mol) + (3 × 16.00 g/mol)
3. Calculate the molar ratio between calcium chloride dihydrate and sodium carbonate using stoichiometry. The balanced chemical equation for the reaction is:
CaCl2·2H2O + Na2CO3 → CaCO3 + 2NaCl + 2H2O
From the equation, we can see that 1 mole of calcium chloride dihydrate reacts with 1 mole of sodium carbonate.
4. Convert the mass of calcium chloride dihydrate into moles:
Moles of CaCl2·2H2O = Mass of CaCl2·2H2O / Molar mass of CaCl2·2H2O
5. Using the molar ratio from step 3, you can determine the moles of sodium carbonate required:
Moles of Na2CO3 = Moles of CaCl2·2H2O
6. Finally, calculate the mass of sodium carbonate required by multiplying the moles from step 5 by the molar mass of sodium carbonate:
Mass of Na2CO3 = Moles of Na2CO3 × Molar mass of Na2CO3
By following these steps, you can find the number of grams of sodium carbonate required for the complete reaction with 1.00g of calcium chloride dihydrate. Remember to use accurate molar masses from reliable sources when calculating.