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?

18.5g Na2CO3

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

the answer is true. hope it .