A solution of 100.00mL of 0.2000mol/L sodium carbonate and 200.00mL of 0.1000mol/L calcium nitrate solutions are mixed together according to the reaction: sodium carbonate + calcium nitrate→ calcium carbonate + sodium nitrate. how would you calculate the mass of calcium carbonate that would precipitate?

A limiting reagent problem.

Na2CO3 + Ca(NO3)2 ==> CaCO3 + 2NaNO3
mol Na2CO3 = M x L = 0.200 x 0.100 = 0.02
mol Ca(NO3)2 = M x L = 0.100 x 0.200 = 0.02 mol.

Calculate how much CaCO3 is formed if we use 0.02 mol Na2CO3 and all the Ca(NO3)2 needed. That will be
0.02 mol Na2CO3 x (1 mol CaCO3/1 mol Na2CO3) = 0.02 mol CaCO3 formed.

Next calculate how much CaCO3 is formed if we use 0.02 mol Ca(NO3)2 and all the Na2CO3 needed. That will be
0.02 mol Ca(NO3)2 x (1 mol CaCO3/1 mol Ca(NO3)2) = 0.02 mol.

Since the number is the same (0.02 mol CaCO3), neither Ca(NO3)2 nor Na2CO3 is the limiting reagent and 0.02 mol CaCO3 is formed.

g CaCO3 = mols x molar mass = ?

1.201g

This answer seems to be correct but you should be explaining in details please

To calculate the mass of calcium carbonate that would precipitate, you need to determine the limiting reactant first. The limiting reactant is the reactant that will be completely used up in the reaction and determines the maximum amount of product that can be formed.

Step 1: Calculate the number of moles of sodium carbonate and calcium nitrate.
Number of moles = concentration (mol/L) × volume (L)
For sodium carbonate:
Number of moles = 0.2000 mol/L × 0.1000 L = 0.0200 mol
For calcium nitrate:
Number of moles = 0.1000 mol/L × 0.2000 L = 0.0200 mol

Step 2: Determine the mole ratio between calcium carbonate and calcium nitrate from the balanced equation.
From the balanced equation, we can see that 1 mole of calcium nitrate reacts with 1 mole of calcium carbonate. Therefore, the mole ratio is 1:1.

Step 3: Identify the limiting reactant.
Compare the number of moles of sodium carbonate and calcium nitrate. Both have the same number of moles, so they are in a 1:1 ratio. Therefore, both reactants are in stoichiometric excess, and none of them is limiting.

Step 4: Calculate the mass of calcium carbonate formed.
From step 3, we determined that both reactants are in stoichiometric excess, meaning they will fully react. The reaction produces an equal amount of calcium carbonate and sodium nitrate. Therefore, the mass of calcium carbonate formed will be the same as the mass of sodium carbonate reacted.

Mass = number of moles × molar mass
Mass of calcium carbonate = 0.0200 mol × (40.08 g/mol)
Mass of calcium carbonate = 0.802 g

So, the mass of calcium carbonate that would precipitate is 0.802 grams.