If a solution containing 117.63 g of silver chlorate is allowed to react completely with a solution containing 10.23 g of lithium bromide, how many grams of solid precipitate will be formed

18.8 g

Explanation:

The equation of the reaction is;

AgClO3(aq) + LiBr(aq)>LiClO3(aq) + AgBr(s)

Number of moles of AgClO3 = 117.63 g/191.32 g/mol = 0.6 moles

Number of moles of LiBr = 10.23 g/86.845 g/mol  = 0.1 moles

Since the molar ratio is 1:1, LiBr is the limiting reactant

Molar mass of solid AgBr = 187.77 g/mol

Mass of precipitate formed = 0.1 moles * 187.77 g/mol

Mass of precipitate formed = 18.8 g

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