Map dth General Chemistr University Science Books presented by Sapling Lea Donal

Question

Map dth General Chemistr University Science Books presented by Sapling Lea Donald McQuarrie. Peter A Rock .Ethan Gallogly If a solution containing 113.81 g of silver chlorate is allowed to react completely with a solution containing 12.53 g of lithium chloride, how many grams of solid precipitate will be formed Number 42.36 How many grams of the reactant in excess will remain after the reaction? Number 1.4 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. Number Number mol mol CI Li Number Ag CIO mol mol

Explanation / Answer

the reaction is given by

AgCl03 + LiCl ---> AgCl (s) + LiCl03

we know that

moles = mass / molar mass

so

moles of AgCl03 = 113.81 / 191.319

moles of AgCl03 = 0.60

now

moles of LiCl = 12.53 / 42.394

moles of liCl = 0.3

now

consider the reaction

AgCl03 + LiCl ---> AgCl (s) + LiCl03

moles of AgCl03 reacted = moles of LiCl added = 0.3

so

moles of AgCl03 remainign = 0.6 - 0.3 = 0.3

moles of LiCl remaining = 0.3 - 0.3 = 0

moles of AgCl formed = 0.3

moles of LiCl03 formed = 0.3

now

the precipitate here is AgCl

so

mass of AgCl = moles x molar mass

so

mass of AgCl formed = 0.3 x 143.32

mass of AgCl formed = 42.36

so

42.36 grams of precipitate is formed

now

excess mass of AgCl03 = 0.3 x 191.319

excess mass of AgCl03 = 57.4

so

57.4 grams of excess reactant remain after the reaction

now

Li+ moles = 0.3

chloride moles = 0

Ag+ moles = 0.3

Cl03- moles = 0.6

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