Answer:
6 Percent Composition. 1. Molar Mass of KBr K = 1(39.10) = 39.10 Br =1(79.90) =79.90 MM = 119.0 79.90 g 119.0 g = 0.6714 3. 0.6714 x 50.0g = 33.6 g Br 2.
The mass of bromine that is present in 50.0 g of potassium bromide is 80 grams.
Relationship between moles and mass will be represented by the below equation:
n = W/M, where
W = given mass
M = molar mass
First we convert the mass of KBr into moles to calculate the moles of bromine through the stoichiometry as:
Moles of KBr = 50g / 119g/mol = 0.42 mol
Dissociation reaction will be represented as:
KBr → K⁺ + Br⁻
From the stoichiometry of the reaction it is clear that 1 mole of KBr produces 1 mole of Bromide, means 1 mole of bromine is present in 1 mole of KBr.
Now mass of Br will be calculated by using the above equation as:
Mass of Br = (1mol)(80g/mol) = 80g
Hence required mass of bromine is 80 grams.
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