Answer:
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
If in a solution reaction between 117.63 g of silver chlorate and 10.23 g of lithium bromide takes place then, 18.8 grams of solid precipitate will be formed.
In any chemical reaction limiting reagent is that reagent whose moles are present in less quantity as compared to the other reagent, is responsible for the formation of product.
Given chemical reaction is:
AgClO₃(aq) + LiBr(aq) → LiClO₃(aq) + AgBr(s)
Moles (n) will ce calculated as:
n = W/M, where
W = given mass
M = molar mass
Moles of AgClO₃ = 117.63g / 191.32g/mol = 0.6 moles
Moles of LiBr = 10.23g / 86.845g/mol = 0.1 moles
In this reaction LiBr is the limiting reagent and formation of product depends on this.
From the stoichiometry of the reaction it is clear that:
0.1 mole of LiBr = produce 0.1 mole of AgBr
Mass of AgBr = (0.1mol)(187.77g/mol) = 18.8 g
Hence mass of AgBr is 18.8 g.
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