Question

Suppose that you chose sodium carbonate to precipitate the chromium ions from a solution of chromium (III) chloride. Write and balance the equation of this double-displacement reaction.

If the solution has a volume of 520 mL and the concentration of chromium (III) chloride is 0.224 M, how many grams of sodium carbonate should you add to the solution to precipitate out all the chromium ions?

Answer 1

The precipitate the chromium ions from a solution of chromium (III) chloride using sodium carbonate, the balanced equation for the reaction is 2Na⁺ (aq) + CO₃(aq) + Cr₃⁺(aq) + 3Cl⁻(aq) → CrCO₃(s) + 2Na+(aq) + 3Cl⁻(aq). The number of grams of sodium carbonate needed to precipitate out all the chromium ions is 12.3457 grams.

Step-by-step explanation:

To precipitate the chromium ions from a solution of chromium (III) chloride using sodium carbonate, we can write the balanced equation for the reaction:

2Na⁺ (aq) + CO₃(aq) + Cr₃⁺(aq) + 3Cl⁻(aq) → CrCO₃(s) + 2Na+(aq) + 3Cl⁻(aq)

This reaction shows that two sodium ions, one carbonate ion, one chromium ion, and three chloride ions react to form one chromium carbonate precipitate, two sodium ions, and three chloride ions.

To calculate the number of grams of sodium carbonate needed to precipitate out all the chromium ions, we need to use the stoichiometry of the reaction.

The molar ratio between chromium ions and sodium carbonate is 1:1, which means that one mole of chromium ions will react with one mole of sodium carbonate.

The concentration of chromium (III) chloride is 0.224 M, which means that there are 0.224 moles of chromium ions in 1 liter of the solution.

The volume of the solution is 520 mL, which is 0.52 L.

Therefore, the number of moles of chromium ions in the solution is 0.224 mol/L * 0.52 L = 0.11648 moles.

Since the molar ratio between chromium ions and sodium carbonate is 1:1, the number of moles of sodium carbonate needed to react with all the chromium ions is also 0.11648 moles.

Using the molar mass of sodium carbonate (105.9888 g/mol), we can calculate the mass of sodium carbonate needed: 0.11648 moles * 105.9888 g/mol = 12.3457 grams.

Therefore the number of grams of sodium carbonate is 12.3457 grams.

Answer 2

18.5g Na2CO3

Step-by-step explanation:

Chromium (III) chloride, CrCl3, reacts with Na2CO3 as follows:

2CrCl3 + 3Na2CO3 → Cr2(CO3)3(s) + 6NaCl

Where 2 moles of CrCl3 react with 3 moles of Na2CO3 to produce 1 mole of Cr2(CO3)3 -The precipitate-

To solve this question we need to find the moles of CrCl3 added. With the chemical equation we can find the moles of Na2CO3 and its mass as follows:

Moles CrCl3:

520mL = 0.520L * (0.224mol/L) = 0.116 moles CrCl3

Moles Na2CO3:

0.116 moles CrCl3 * (3 mol Na2CO3 / 2mol CrCl3) = 0.175 moles Na2CO3

Mass Na2CO3 -Molar mass: 105.99g/mol-

0.175 moles Na2CO3 * (105.99g/mol) = 18.5g Na2CO3