Mercury and Bromine will react with each other according to the reaction: Hg (l)
ID: 615081 • Letter: M
Question
Mercury and Bromine will react with each other according to the reaction: Hg (l) + Br2 (l) ? HgBr2 (s) What mass of mercury (II) bromide can be produced from the reaction of 10.0 g Hg and 9.00 g Br2? How much of the excess reagent remains at the end of the experiment?Explanation / Answer
Hg + Br2 ----->HgBr2 a)calculate moles of each reactant and see which one is less, that would be the limiting reactant because here is 1:1 molar mass of Hg is 200 and Br2 is 160 mole=gr/molar mass Hg moles=10/200 = 0.05 Br2 moles=9/160 = 0.056 As Hg has less moles it's the limiting reactant. The limiting reactants in complete equations will end so 0.05 mole Hg reacts with 0.05 mole Br2 and produces 0.05 mole HgBr2 (1:1:1). 0.05 mole Br2 reacted ,so 0.056 - 0.05=0.006 mole of Br2 is left, change to gr 0.006 mole Br2 = gr / 160 , gr Br2=0.96 gr is left 0.05 mole HgBr2 produced , change it to gr:(molar mass of HgBr2 is 360) 0.05=gr/360, grHgBr2=18 gr is produced b) It's just like the previous problem but you need to change them to mole by their density: 5 ml Hg x (13.6 gr Hg/ 1ml Hg) x (1 mole Hg/ 200 gr Hg) = 0.034 mole Hg 5 ml Br2 x (3.1 gr Br2 /1 ml Br2) x (1 mole Br2 /160 gr Br2) = 0.096 mole Br2 so the limiting reactant is Hg again . 0.034 mole Hg reacts with 0.034 mole Br2 and produces 0.034 mole HgBr2 (1:1:1) , change 0.034 mole HgBr2 to gr: 0.034 mole HgBr2 = gr / 360 gr HgBr2 = 12.24 gr HgBr2 is produced