How many grams of sodium carbonate is required to have a complete reaction with 1.00g of calcium chloride dihydrate? I can't figure this out...help!
Assistance needed.
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I know...I realized that right after I accidentally posted - that is why there is the CORRECT post and subject right after this one called Stoichiometry. THANKS FOR THE HEADS UP, THOUGH.
rucka rucka shut up
To determine the number of grams of sodium carbonate required for a complete reaction with calcium chloride dihydrate, we need to use stoichiometry.
Step 1: Write the balanced chemical equation:
CaCl2·2H2O + Na2CO3 → CaCO3 + 2NaCl + 2H2O
This equation shows that 1 mole of calcium chloride dihydrate (CaCl2·2H2O) reacts with 1 mole of sodium carbonate (Na2CO3) to produce 1 mole of calcium carbonate (CaCO3), 2 moles of sodium chloride (2NaCl), and 2 moles of water (2H2O).
Step 2: Calculate the molar mass:
- Calcium chloride dihydrate (CaCl2·2H2O):
- Molar mass of Ca = 40.08 g/mol
- Molar mass of Cl2 = 35.45 g/mol x 2 = 70.90 g/mol
- Molar mass of H2O = 18.02 g/mol x 2 = 36.04 g/mol
- Total molar mass = 40.08 + 70.90 + 36.04 = 147.02 g/mol
- Sodium carbonate (Na2CO3):
- Molar mass of Na2 = 22.99 g/mol x 2 = 45.98 g/mol
- Molar mass of C = 12.01 g/mol
- Molar mass of O3 = 16.00 g/mol x 3 = 48.00 g/mol
- Total molar mass = 45.98 + 12.01 + 48.00 = 105.99 g/mol
Step 3: Calculate the moles of calcium chloride dihydrate:
- Given mass of CaCl2·2H2O = 1.00 g
- Moles = mass / molar mass
- Moles = 1.00 g / 147.02 g/mol ≈ 0.0068 mol
Step 4: Determine the moles of sodium carbonate:
- According to the balanced equation, the mole ratio between calcium chloride dihydrate and sodium carbonate is 1:1.
- Therefore, moles of Na2CO3 = 0.0068 mol
Step 5: Calculate the mass of sodium carbonate:
- Mass = moles x molar mass
- Mass = 0.0068 mol x 105.99 g/mol ≈ 0.719 g
Therefore, approximately 0.719 grams of sodium carbonate is required for a complete reaction with 1.00 gram of calcium chloride dihydrate.