Answered

Westonci.ca connects you with experts who provide insightful answers to your questions. Join us today and start learning! Experience the convenience of finding accurate answers to your questions from knowledgeable professionals on our platform. Get quick and reliable solutions to your questions from a community of experienced experts on our platform.

three moles of sodium carbonate are mixed with two moles of lead nitrate in aqueous solution, leading to formation of a solid precipitate. how many moles of spectator ions remain in solution, assuming 100% yield of the precipitate?

Sagot :

pstnonsonjoku

There are 4 moles of spectator ions that remain in solution.

The equation of the reaction is;

Na2CO3(aq) + Pb(NO3)2(aq) -------> PbCO3(s) + 2NaNO3(aq)

We have to determine the limiting reactant. This is the reactant that yields the least amount of product. Note that the spectator ions are Na^+ and NO3^- that form NaNO3.

For Na2CO3

1 mole of Na2CO3 yields 2 moles of NaNO3

3 moles of Na2CO3 yields  3 × 2/1 = 6 moles of NaNO3

For Pb(NO3)2

1 mole of Pb(NO3)2 yields 2 moles of NaNO3

2 moles of Pb(NO3)2 yields 2 × 2/1 = 4 moles of NaNO3

We can see that Pb(NO3)2 is the limiting reactant.

Since [NaNO3] = [Na^+] = [NO3^-], it follows that there are 4 moles of spectator ions that remain in solution.

Learn more: https://brainly.com/question/22885959

We appreciate your visit. Hopefully, the answers you found were beneficial. Don't hesitate to come back for more information. We hope you found this helpful. Feel free to come back anytime for more accurate answers and updated information. Westonci.ca is committed to providing accurate answers. Come back soon for more trustworthy information.