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How many moles of KBr will be produced from 4.79 moles of BaBr2?

Anonymous

∙ 9y ago

Updated: 6/4/2024

14.17 mol BaBr2 has 2*14.17 mol Br in it, so 28.34 mol KBr can be produced (also 28.34 mol K is needed)

Wiki User

∙ 9y ago

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Aluminum will react with bromine to form aluminum bromide Al2Br6 What mass of bromine g is needed to form 1.42 mol of Al2Br6?

The gram formula unit or molar mass for aluminum bromide is 533.38.* Therefore, 1.42 moles has a mass of 757.4 grams. The mass of 6 moles of bromine atoms is 479.42. Therefore, the mass fraction of bromine in aluminum bromide is 479.42/757.4 or 0.633, and the mass in grams of bromine required to form 1.42 moles of aluminum bromide is 0.633 X 757.4 or 479 grams, to the justified number of significant digits (limited by the precision given for the number of moles.) ___________________ *This is equal to the sum of (2 times the gram atomic mass of aluminum) and (6 times the gram atomic mass of bromine).

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