rigrace21
Answered

Discover a world of knowledge at Westonci.ca, where experts and enthusiasts come together to answer your questions. Get immediate and reliable solutions to your questions from a knowledgeable community of professionals on our platform. Our platform offers a seamless experience for finding reliable answers from a network of knowledgeable professionals.

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?

Sagot :

jufsanabriasa

Answer:

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

We hope our answers were helpful. Return anytime for more information and answers to any other questions you may have. We hope you found this helpful. Feel free to come back anytime for more accurate answers and updated information. We're glad you chose Westonci.ca. Revisit us for updated answers from our knowledgeable team.