1. To write the balanced equation for the reaction between aqueous Pb(CIO3)2 and aqueous NaI, we need to know the formulas of the compounds involved.
The formula for lead(II) chlorate is Pb(CIO3)2, and the formula for sodium iodide is NaI.
The balanced equation is:
Pb(CIO3)2(aq) + 2NaI(aq) → PbI2(s) + 2NaCIO3(aq)
2. To find out how many grams of solid precipitate will be formed when 86.044 g of mercury(II) chlorate reacts completely with 15.4888 g of sodium sulfide, we need to know the balanced equation for the reaction.
The balanced equation for the reaction between mercury(II) chlorate and sodium sulfide is:
Hg(CIO3)2 + Na2S → HgS + 2NaCIO3
From the equation, we can see that the stoichiometry between Hg(CIO3)2 and HgS is 1:1. This means that for every 1 mole of Hg(CIO3)2, 1 mole of HgS is formed.
First, we need to calculate the number of moles of Hg(CIO3)2 and Na2S:
- Moles of Hg(CIO3)2 = mass of Hg(CIO3)2 / molar mass of Hg(CIO3)2
- Moles of Na2S = mass of Na2S / molar mass of Na2S
Then, we compare the moles of Hg(CIO3)2 and Na2S to see which one is the limiting reactant (i.e., the reactant that will be completely consumed). The reactant with the smaller number of moles is the limiting reactant.
Using the balanced equation, we can determine the number of moles of HgS that will be formed. Since the stoichiometry between Hg(CIO3)2 and HgS is 1:1, the number of moles of HgS formed will be equal to the moles of the limiting reactant.
Finally, we calculate the mass of the solid precipitate using the molar mass of HgS:
- Mass of HgS precipitate = moles of HgS formed x molar mass of HgS
3. To determine how many grams of the excess reactant will remain after the reaction, we first need to identify the excess reactant. The excess reactant is the reactant that remains after the limiting reactant is completely consumed.
Using the balanced equation and the moles of the limiting reactant, we can calculate the moles of each reactant consumed. From there, we can find the moles of the excess reactant remaining.
Finally, we calculate the mass of the excess reactant remaining using the moles of the excess reactant and the molar mass of the excess reactant.
Please note that to provide precise calculations, we need the molar masses of the compounds involved (Hg(CIO3)2, Na2S, HgS, etc.).