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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?

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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

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