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.

Question
Asked Sep 19, 2019
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Liquid mercury and bromine gas will react under appropriate conditions to produce solid mercury (II) bromide.

a. What is the maximum mass of HgBrthat 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.

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Expert Answer

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Step 1

(a)

Determine the limiting reagent:

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
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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

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Step 2

Determine the maximum amount of ...

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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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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