Answer:
The mass of the white solid calcium carbonate that forms is 0.300 g.
Step-by-step explanation:
To calculate the mass of the white solid calcium carbonate that forms, we first need to determine the limiting reagent in the reaction between calcium nitrate and sodium carbonate. The balanced chemical equation for the reaction is:
Ca(NO3)2(aq) + Na2CO3(aq) → CaCO3(s) + 2NaNO3(aq)
From the equation, we can see that one mole of calcium nitrate reacts with one mole of sodium carbonate to produce one mole of calcium carbonate. Therefore, the limiting reagent is the one that produces the least amount of calcium carbonate.
To determine the limiting reagent, we need to calculate the moles of calcium nitrate and sodium carbonate used in the reaction:
Moles of calcium nitrate = volume of solution (L) x concentration (mol/L) = 25.0 L x 0.100 mol/L = 2.50 mol
Moles of sodium carbonate = volume of solution (L) x concentration (mol/L) = 0.0200 L x 0.150 mol/L = 0.00300 mol
Since the moles of sodium carbonate are much smaller than the moles of calcium nitrate, sodium carbonate is the limiting reagent.
The balanced chemical equation tells us that one mole of calcium carbonate is produced for every mole of sodium carbonate used. Therefore, the moles of calcium carbonate produced are also equal to 0.00300 mol.
Finally, we can calculate the mass of calcium carbonate produced using the molar mass of calcium carbonate:
Mass = moles x molar mass = 0.00300 mol x 100.1 g/mol = 0.300 g
Therefore, the mass of the white solid calcium carbonate that forms is 0.300 g.