Question

If a solution containing 15.63 g of lead(II) chlorate is allowed to react completely with a solution containing 5.102 g of sodium sulfate, how many grams of solid precipitate will be formed?

How many grams of the reactant in excess will remain after the reaction?

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+

ClO3-

SO4^2-

Na+

Answer 1