# A solution containing sodium carbonate weighing 6.36 g was added to a solution containing calcium

**A solution containing sodium carbonate weighing 6.36 g was added to a solution containing calcium nitrate weighing 8.2 g. Calculate the mass of the resulting precipitate.**

Let’s execute the solution:

1. Let’s write the equation:

Ca (NO3) 2 + Na2CO3 = CaCO3 + 2NaNO3 – ion exchange, obtained calcium carbonate in the sediment;

2. Let’s make calculations using the formulas:

M Ca (NO3) 2 = 164 g / mol.

M (Na2CO3) = 105.8 g / mol.

M (CaCO3) = 100 g / mol.

3. Determine Y (amount) of starting substances:

Y Ca (NO3) 2 = m / M = 8.2 / 164 = 0.05 mol (deficient substance).

Y (CaCO3) = 0.05 mol since the amount of substances according to the equation is 1 mol.

Y (Na2CO3) = m / M = 6.36 / 105.8 = 0.06 mol (substance in excess).

Calculations are made for the substance in deficiency.

4. Find the mass of the product:

m (CaCO3) = Y * M = 0.05 * 100 = 5 g.

Answer: The mass of calcium carbonate is 5 g.