# 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.

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