If a solution containing 54.2154.21 g of lead(II) chlorate is allowed to react completely with a solution containing 8.5648.564 g of sodium sulfide, how many grams of solid precipitate will be formed? mass of solid precipitate: gg How many grams of the reactant in excess will remain after the reaction? mass of excess reactant: gg 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+=Pb2+= molmol ClO–3=ClO3–= molmol Na+=Na+= molmol S2−S2− mol
If a solution containing 54.2154.21 g of lead(II) chlorate is allowed to react completely with a solution containing 8.5648.564 g of sodium sulfide, how many grams of solid precipitate will be formed? mass of solid precipitate: gg How many grams of the reactant in excess will remain after the reaction? mass of excess reactant: gg 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+=Pb2+= molmol ClO–3=ClO3–= molmol Na+=Na+= molmol S2−S2− mol
Chapter3: Stoichiometry
Section: Chapter Questions
Problem 159CWP: Sulfur dioxide gas reacts with sodium hydroxide to form sodium sulfite and water. The unbalanced...
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If a solution containing 54.2154.21 g of lead(II) chlorate is allowed to react completely with a solution containing 8.5648.564 g of sodium sulfide, how many grams of solid precipitate will be formed?
mass of solid precipitate:
gg
How many grams of the reactant in excess will remain after the reaction?
mass of excess reactant:
gg
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+=Pb2+=
molmol
ClO–3=ClO3–=
molmol
Na+=Na+=
molmol
S2−S2−
mol
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