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.
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?
1 answer
1 answer