If a solution containing 61.28 g of lead(II) chlorate is allowed to react completely with a solution containing 8.564 g of sodium sulfide, how many grams of solid precipitate will be formed? mass of solid precipitate: g How many grams of the reactant in excess will remain after the reaction? mass of excess reactant:

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ChapterU4: Toxins: Stoichiometry, Solution Chemistry, And Acids And Bases
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If a solution containing 61.28 g of lead(II) chlorate is allowed to react completely with a solution containing 8.564 g of
sodium sulfide, how many grams of solid precipitate will be formed?
mass of solid precipitate:
g
How many grams of the reactant in excess will remain after the reaction?
mass of excess reactant:
g
Transcribed Image Text:If a solution containing 61.28 g of lead(II) chlorate is allowed to react completely with a solution containing 8.564 g of sodium sulfide, how many grams of solid precipitate will be formed? mass of solid precipitate: g How many grams of the reactant in excess will remain after the reaction? mass of excess reactant: g
Assuming complete precipitation, how many moles of each ion remain in solution? If an ion is no longer in solution, enter a
zero (0) for the number of moles.
Pb2+:
mol
C1O3:
mol
Nat:
mol
S²-.
mol
Transcribed Image Text:Assuming complete precipitation, how many moles of each ion remain in solution? If an ion is no longer in solution, enter a zero (0) for the number of moles. Pb2+: mol C1O3: mol Nat: mol S²-. mol
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