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Social ScienceLiquid mercury and bromine gas will react under appropriate conditions to produce solid mercury (II) bromide.a. What is the maximum mass of HgBr2 that can be produced from the reaction of 10.0 g Hg and 9.00 g Br2?b. Determine the remaining mass of each reactant available upon conclusion of the reaction.
Liquid mercury and bromine gas will react under appropriate conditions to produce solid mercury (II) bromide.
a. What is the maximum mass of HgBr2 that can be produced from the reaction of 10.0 g Hg and 9.00 g Br2?
b. Determine the remaining mass of each reactant available upon conclusion of the reaction.
Expert Answer
(a)
Determine the limiting reagent:
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Given reaction: Hg(1) + Br3 (g) HgBr, (s) 10.0 g Mass No. of moles of Hg = 0.0498 mol 1 Molar mass 200.59 g/mol 9.0 g No. of moles of Br = 0.0563 mol 159.81 g/mol One mole of Hg and one mol of Br, reacts to give one mol HgBr No. of moles ofHg < No. of moles of Br Hence, limiting reagent: Hg
Determine the maximum amount of ...
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mol HgBr 1 mol Hg 0.0498 mol Hgx 1 mol HgBr 1 mol Hg = 0.0498 mol HgBr No. of moles of HgBr, formed Mass of HgBr, fomed = No. of moles x Molar mass 0.0498 molx 360.40 g/mol =17.95 g Therefore, 17.95 g of HgBr, is produced
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