Question

For the following reaction : CaCl₂(aq) + 2NaHCO₃(aq) -> CaCO₃(s) + H₂O(l) + CO₂(g) + 2NaCl(aq)

molar mass of CaCl₂ = 110.98 g/mol
molar mass of 2NaHCO₃ = 84.007 g/mol
molar mass of CaCO₃ = 100.09 g/mol

The CaCl₂ used in this experiment exist as a dihydrate, CaCl₂ . 2H₂O. If 1.00g of CaCl₂. 2H₂O reacted with an excess amount of NaHCO₃, how many grams of solid CaCO₃ should precipitate out?

Answer 1

Around 0.68 grams of solid CaCO3 will precipitate when 1.00g of CaCl2· 2H2O is reacted with an excess amount of NaHCO3, after calculating the molarity based on molar masses and using stoichiometry from the balanced reaction.

Step-by-step explanation:

When calculating the amount of solid CaCO3 that will precipitate from the reaction of 1.00g of CaCl2· 2H2O with an excess amount of NaHCO3, we must first take into account the molar mass of CaCl2· 2H2O. The molar mass of CaCl2 is 110.98 g/mol, and the molar mass of water (H2O) is approximately 18.015 g/mol. Therefore, the molar mass of CaCl2· 2H2O is the sum of these two values: 110.98 g/mol + 2(18.015 g/mol) = 147.01 g/mol.

To determine the number of moles of CaCl2· 2H2O in 1.00 g, we divide the mass of the sample by the molar mass: 1.00 g ÷ 147.01 g/mol = 0.0068 moles. From the balanced chemical equation, we see that one mole of CaCl2 produces one mole of solid CaCO3. Therefore, 0.0068 moles of CaCl2 should produce 0.0068 moles of CaCO3.

Finally, we calculate the mass of solid CaCO3 by multiplying the moles of CaCO3 by its molar mass: 0.0068 moles × 100.09 g/mol = 0.68161 grams of CaCO3. We can estimate that approximately 0.68 grams of CaCO3 will precipitate from the reaction.