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
If 1.00g of NaHCO₃ reacted with an excess amount of CaCl₂, how many grams of solid CaCO₃ should precipitate out?

Answer 1

When 1.00 g of NaHCO₃ reacts with an excess amount of CaCl₂, approximately 0.5957 grams of CaCO₃ should precipitate out after calculating moles of reactant and product and converting this to grams using the molar mass of CaCO₃.

Step-by-step explanation:

Calculation of mass of CaCO₃ Precipitate:

To calculate the mass of CaCO₃ precipitated when 1.00 g of NaHCO₃ reacts with an excess of CaCl₂, we first determine the number of moles of NaHCO₃:

The molar mass of 2NaHCO₃ = 84.007 g/mol (for 2 moles, so for 1 mole it would be half)

1.00 g NaHCO₃ * (1 mol/84.007 g) = 0.011902 moles of NaHCO₃

From the balanced chemical equation, 2 moles of NaHCO₃ yields 1 mole of CaCO₃. Therefore, 0.011902 moles of NaHCO₃ will yield:

0.011902 moles NaHCO₃ * (1 mol CaCO₃ / 2 moles NaHCO₃) = 0.005951 moles of CaCO₃

Now, to find the mass of CaCO₃ precipitated:

0.005951 moles CaCO₃ * 100.09 g/mol (molar mass of CaCO₃) = 0.5957 g of CaCO₃

Therefore, when 1.00 g of NaHCO₃ is reacted with an excess amount of CaCl₂, approximately 0.5957 grams of CaCO₃ should precipitate out.