write the balanced equation for the reactions of aquesous pb(CIO3)2 with aqueous Nal. include phases.
2.If a solution containing 86.044 g of mercury(II) chlorate is allowed to react completely cointaing 15.4888 g of sodium sulfide, how many grams of solid precipate will be formed?
b. how many grams of the reactant in excess will remain after the reaction?
To write the balanced equation for the reaction between aqueous Pb(ClO3)2 and aqueous NaI:
1. Identify the formulas of the compounds involved:
- Pb(ClO3)2 represents lead(II) chlorate.
- NaI represents sodium iodide.
2. Write the chemical equation by swapping the ions between the compounds:
- Pb(ClO3)2 (aq) + 2 NaI (aq) → PbI2 (s) + 2 NaClO3 (aq)
In the balanced equation, the coefficients in front of each compound represent the stoichiometric ratio of the reactants and products. Hence, 1 mole of Pb(ClO3)2 reacts with 2 moles of NaI to form 1 mole of PbI2 and 2 moles of NaClO3.
Now, addressing the second question:
To find the amount of solid precipitate formed and the excess reactant remaining, we need to calculate the limiting reagent first.
1. Calculate the molar mass of each compound:
- Mercury(II) chlorate (Hg(ClO3)2) = 271.5 g/mol
- Sodium sulfide (Na2S) = 78.04 g/mol
2. Convert the given masses of the reactants to moles:
- Moles of Hg(ClO3)2 = 86.044 g / 271.5 g/mol = 0.3172 mol
- Moles of Na2S = 15.4888 g / 78.04 g/mol = 0.1984 mol
3. Determine the stoichiometric ratio between the reactants in the balanced equation:
From the balanced equation: 1 mole of Hg(ClO3)2 reacts with 1 mole of Na2S.
4. Identify the limiting reagent:
Compare the moles of each reactant to the stoichiometric ratio.
- Hg(ClO3)2: 0.3172 mol
- Na2S: 0.1984 mol
Since the stoichiometric ratio is 1:1, we can see that Na2S is the limiting reagent (the smaller value).
5. Calculate the mass of solid precipitate formed (PbS) using the stoichiometry from the balanced equation:
From the balanced equation: 1 mole of Na2S forms 1 mole of PbS.
- Moles of PbS = 0.1984 mol
- Mass of PbS = Moles of PbS × Molar mass of PbS
6. Calculate the mass of the excess reactant remaining:
Since Na2S is the limiting reagent, we need to find the excess amount of Hg(ClO3)2.
- Moles of Hg(ClO3)2 in excess = Moles of Hg(ClO3)2 - Moles of Na2S
- Mass of Hg(ClO3)2 in excess = Moles of Hg(ClO3)2 in excess × Molar mass of Hg(ClO3)2
By following these steps, you can calculate the mass of the solid precipitate formed and the mass of the excess reactant remaining in the reaction.
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.).
Pb(ClO3)2(aq) + 2NaI(aq)==> PbI2 + 2NaClO3
2.This is a limiting reagent problem. I know that because amounts are given for BOTH reactants.
Hg(ClO3)2 + 2Na2S ==> HgS(s) + 2NaClO3
Convert 86.044 g Hg(ClO3)2 to mols. mols = gram/molar mass.
Do the same for Na2S .
Using the coefficients in the balanced equation, convert mols Hg(ClO3)2 to mols HgS.
Do the same for mols Na2S to mols HgS.
It is likely that these two values will not be the same which means one of them is not right; the correct answer in limiting reagent problems is ALWAYS the smaller value and the reagent responsible for the smaller value will be the limiting reagent(LR). The other reagent is called "other reagent" or "excess reagent (ER)" or "non-limiting" or some such term. You can convert mols HgS to grams by g = mols x molar mass.
Now that you have identified the LR and ER, use the coefficients as above to convert mols of the LR used to mols ER used. That will tell you the number of mols ER used. Then initial ER - ER used = mols ER remaining un-reacted. Then g = mols x molar mass.
Post your work if you get stuck or have questions.